How do I calculate the variance of the OLS estimator $\beta_0$, conditional on $x_1, \ldots , x_n$? I know that
$$\hat{\beta_0}=\bar{y}-\hat{\beta_1}\bar{x}$$
and this is how far I got when I calculated the variance:
\begin{align*}
Var(\hat{\beta_0}) &= Var(\bar{y} - \hat{\beta_1}\bar{x}) \\
&= Var((-\bar{x})\hat{\beta_1}+\bar{y}) \\
&= Var((-\bar{x})\hat{\beta_1})+Var(\bar{y}) \\
&= (-\bar{x})^2 Var(\hat{\beta_1}) + 0 \\
&= (\bar{x})^2 Var(\hat{\beta_1}) + 0 \\
&= \frac{\sigma^2 (\bar{x})^2}{\displaystyle\sum\limits_{i=1}^n (x_i - \bar{x})^2}
\end{align*}
but that's far as I got. The final formula I'm trying to calculate is 
\begin{align*}
Var(\hat{\beta_0}) &= \frac{\sigma^2 n^{-1}\displaystyle\sum\limits_{i=1}^n x_i^2}{\displaystyle\sum\limits_{i=1}^n (x_i - \bar{x})^2}
\end{align*}
I'm not sure how to get $$(\bar{x})^2 = \frac{1}{n}\displaystyle\sum\limits_{i=1}^n x_i^2$$ assuming my math is correct up to there.
Is this the right path?
\begin{align}
(\bar{x})^2 &= \left(\frac{1}{n}\displaystyle\sum\limits_{i=1}^n x_i\right)^2 \\
&= \frac{1}{n^2} \left(\displaystyle\sum\limits_{i=1}^n x_i\right)^2
\end{align}
I'm sure it's simple, so the answer can wait for a bit if someone has a hint to push me in the right direction. 
 A: This is a self-study question, so I provide hints that will hopefully help to find the solution, and I'll edit the answer based on your feedbacks/progress.
The parameter estimates that minimize the sum of squares are
\begin{align}
  \hat{\beta}_0 &= \bar{y} - \hat{\beta}_1 \bar{x} , \\
  \hat{\beta}_1 &= \frac{ \sum_{i = 1}^n(x_i - \bar{x})y_i }{ \sum_{i = 1}^n(x_i - \bar{x})^2 } .
\end{align}
To get the variance of $\hat{\beta}_0$, start from its expression and substitute the expression of $\hat{\beta}_1$, and do the algebra
$$
{\rm Var}(\hat{\beta}_0) = {\rm Var} (\bar{Y} - \hat{\beta}_1 \bar{x}) = \ldots
$$
Edit:
We have
\begin{align}
{\rm Var}(\hat{\beta}_0)
 &= {\rm Var} (\bar{Y} - \hat{\beta}_1 \bar{x}) \\
 &= {\rm Var} (\bar{Y}) + (\bar{x})^2 {\rm Var} (\hat{\beta}_1)
    - 2 \bar{x} {\rm Cov} (\bar{Y}, \hat{\beta}_1).
\end{align}
The two variance terms are
$$
{\rm Var} (\bar{Y})
  = {\rm Var} \left(\frac{1}{n} \sum_{i = 1}^n Y_i \right)
  = \frac{1}{n^2} \sum_{i = 1}^n {\rm Var} (Y_i)
  = \frac{\sigma^2}{n},
$$
and
\begin{align}
{\rm Var} (\hat{\beta}_1)
  &= \frac{ 1 }{ \left[\sum_{i = 1}^n(x_i - \bar{x})^2 \right]^2 }
    \sum_{i = 1}^n(x_i - \bar{x})^2 {\rm Var} (Y_i) \\
  &= \frac{ \sigma^2 }{ \sum_{i = 1}^n(x_i - \bar{x})^2  } ,
\end{align}
and the covariance term is
\begin{align}
{\rm Cov} (\bar{Y}, \hat{\beta}_1)
 &= {\rm Cov} \left\{
     \frac{1}{n} \sum_{i = 1}^n Y_i,
     \frac{ \sum_{j = 1}^n(x_j - \bar{x})Y_j }{ \sum_{i = 1}^n(x_i - \bar{x})^2 }
     \right \} \\
 &= \frac{1}{n} \frac{ 1 }{ \sum_{i = 1}^n(x_i - \bar{x})^2 }
    {\rm Cov} \left\{ \sum_{i = 1}^n Y_i, \sum_{j = 1}^n(x_j - \bar{x})Y_j \right\} \\
 &= \frac{ 1 }{ n \sum_{i = 1}^n(x_i - \bar{x})^2 }
    \sum_{i = 1}^n (x_j - \bar{x}) \sum_{j = 1}^n {\rm Cov}(Y_i, Y_j) \\
 &= \frac{ 1 }{ n \sum_{i = 1}^n(x_i - \bar{x})^2 }
    \sum_{i = 1}^n (x_j - \bar{x}) \sigma^2 \\
 &= 0
\end{align}
since $\sum_{i = 1}^n (x_j - \bar{x})=0$.
And since
$$\sum_{i = 1}^n(x_i - \bar{x})^2 
= \sum_{i = 1}^n x_i^2 - 2 \bar{x} \sum_{i = 1}^n x_i 
  + \sum_{i = 1}^n \bar{x}^2
= \sum_{i = 1}^n x_i^2 - n \bar{x}^2,
$$
we have
\begin{align}
{\rm Var}(\hat{\beta}_0)
 &= \frac{\sigma^2}{n} + \frac{ \sigma^2 \bar{x}^2}{ \sum_{i = 1}^n(x_i - \bar{x})^2  } \\
 &= \frac{\sigma^2 }{ n \sum_{i = 1}^n(x_i - \bar{x})^2 }
    \left\{ \sum_{i = 1}^n(x_i - \bar{x})^2 + n \bar{x}^2 \right\} \\
 &= \frac{\sigma^2 \sum_{i = 1}^n x_i^2}{ n \sum_{i = 1}^n(x_i - \bar{x})^2 }.
\end{align}
Edit 2

Why do we have
  ${\rm var} ( \sum_{i = 1}^n Y_i) = \sum_{i = 1}^n {\rm Var} (Y_i) $?

The assumed model is $ Y_i = \beta_0 + \beta_1 X_i + \epsilon_i$, where the $\epsilon_i$ are  independant and identically distributed random variables with ${\rm E}(\epsilon_i) = 0$ and ${\rm var}(\epsilon_i) = \sigma^2$.
Once we have a sample, the $X_i$ are known, the only random terms are the $\epsilon_i$. Recalling that for a random variable $Z$ and a constant $a$, we have ${\rm var}(a+Z) = {\rm var}(Z)$. Thus,
\begin{align}
{\rm var} \left( \sum_{i = 1}^n Y_i \right)
&= {\rm var} \left( \sum_{i = 1}^n \beta_0 + \beta_1 X_i + \epsilon_i \right)\\
&= {\rm var} \left( \sum_{i = 1}^n \epsilon_i \right)
= \sum_{i = 1}^n \sum_{j = 1}^n  {\rm cov} (\epsilon_i, \epsilon_j)\\
&= \sum_{i = 1}^n  {\rm cov} (\epsilon_i, \epsilon_i)
= \sum_{i = 1}^n  {\rm var} (\epsilon_i)\\
&= \sum_{i = 1}^n  {\rm var} (\beta_0 + \beta_1 X_i + \epsilon_i)
= \sum_{i = 1}^n  {\rm var} (Y_i).\\
\end{align}
The 4th equality holds as ${\rm cov} (\epsilon_i, \epsilon_j) = 0$ for $i \neq j$ by the independence of the $\epsilon_i$.
A: I got it! Well, with help. I found the part of the book that gives steps to work through when proving the $Var \left( \hat{\beta}_0 \right)$ formula (thankfully it doesn't actually work them out, otherwise I'd be tempted to not actually do the proof). I proved each separate step, and I think it worked. 
I'm using the book's notation, which is:
$$
SST_x = \displaystyle\sum\limits_{i=1}^n (x_i - \bar{x})^2,
$$
and $u_i$ is the error term.
1) Show that $\hat{\beta}_1$ can be written as $\hat{\beta}_1 = \beta_1 + \displaystyle\sum\limits_{i=1}^n w_i u_i$ where $w_i = \frac{d_i}{SST_x}$ and $d_i = x_i - \bar{x}$.
This was easy because we know that
\begin{align}
\hat{\beta}_1 &= \beta_1 + \frac{\displaystyle\sum\limits_{i=1}^n (x_i - \bar{x}) u_i}{SST_x} \\
&= \beta_1 + \displaystyle\sum\limits_{i=1}^n \frac{d_i}{SST_x} u_i \\
&= \beta_1 + \displaystyle\sum\limits_{i=1}^n w_i u_i
\end{align}
2) Use part 1, along with $\displaystyle\sum\limits_{i=1}^n w_i = 0$ to show that $\hat{\beta_1}$ and $\bar{u}$ are uncorrelated, i.e. show that $E[(\hat{\beta_1}-\beta_1) \bar{u}] = 0$. 
\begin{align}
E[(\hat{\beta_1}-\beta_1) \bar{u}] &= E[\bar{u}\displaystyle\sum\limits_{i=1}^n w_i u_i] \\
&=\displaystyle\sum\limits_{i=1}^n  E[w_i \bar{u} u_i] \\
&=\displaystyle\sum\limits_{i=1}^n w_i E[\bar{u} u_i] \\
&= \frac{1}{n}\displaystyle\sum\limits_{i=1}^n w_i E\left(u_i\displaystyle\sum\limits_{j=1}^n u_j\right) \\
&= \frac{1}{n}\displaystyle\sum\limits_{i=1}^n w_i \left[E\left(u_i u_1\right) +\cdots + E(u_i u_j) + \cdots+ E\left(u_i u_n \right)\right] \\
\end{align}
and because the $u$ are i.i.d., $E(u_i u_j) = E(u_i) E(u_j)$ when $ j \neq i$.
When $j = i$, $E(u_i u_j) = E(u_i^2)$, so we have:
\begin{align}
&= \frac{1}{n}\displaystyle\sum\limits_{i=1}^n w_i \left[E(u_i) E(u_1) +\cdots + E(u_i^2) + \cdots + E(u_i) E(u_n)\right] \\
&= \frac{1}{n}\displaystyle\sum\limits_{i=1}^n w_i E(u_i^2) \\
&= \frac{1}{n}\displaystyle\sum\limits_{i=1}^n w_i \left[Var(u_i) + E(u_i) E(u_i)\right] \\
&= \frac{1}{n}\displaystyle\sum\limits_{i=1}^n w_i \sigma^2 \\
&= \frac{\sigma^2}{n}\displaystyle\sum\limits_{i=1}^n w_i \\
&= \frac{\sigma^2}{n \cdot SST_x}\displaystyle\sum\limits_{i=1}^n (x_i - \bar{x}) \\
&= \frac{\sigma^2}{n \cdot SST_x} \left(0\right)
&= 0
\end{align}
3) Show that $\hat{\beta_0}$ can be written as $\hat{\beta_0} = \beta_0 + \bar{u} - \bar{x}(\hat{\beta_1} - \beta_1)$. This seemed pretty easy too:
\begin{align}
\hat{\beta_0} &= \bar{y} - \hat{\beta_1} \bar{x} \\
&= (\beta_0 + \beta_1 \bar{x} + \bar{u}) - \hat{\beta_1} \bar{x} \\
&= \beta_0 + \bar{u} - \bar{x}(\hat{\beta_1} - \beta_1).
\end{align}
4) Use parts 2 and 3 to show that $Var(\hat{\beta_0}) = \frac{\sigma^2}{n} + \frac{\sigma^2 (\bar{x}) ^2} {SST_x}$:
\begin{align}
Var(\hat{\beta_0}) &= Var(\beta_0 + \bar{u} - \bar{x}(\hat{\beta_1} - \beta_1)) \\
&= Var(\bar{u}) + (-\bar{x})^2 Var(\hat{\beta_1} - \beta_1) \\
&= \frac{\sigma^2}{n} + (\bar{x})^2 Var(\hat{\beta_1}) \\
&= \frac{\sigma^2}{n} + \frac{\sigma^2 (\bar{x}) ^2} {SST_x}.
\end{align}
I believe this all works because since we provided that $\bar{u}$ and $\hat{\beta_1} - \beta_1$ are uncorrelated, the covariance between them is zero, so the variance of the sum is the sum of the variance. $\beta_0$ is just a constant, so it drops out, as does $\beta_1$ later in the calculations.
5) Use algebra and the fact that $\frac{SST_x}{n} = \frac{1}{n} \displaystyle\sum\limits_{i=1}^n x_i^2 - (\bar{x})^2$:
\begin{align}
Var(\hat{\beta_0}) &= \frac{\sigma^2}{n} + \frac{\sigma^2 (\bar{x}) ^2} {SST_x} \\
&= \frac{\sigma^2 SST_x}{SST_x n} + \frac{\sigma^2 (\bar{x})^2}{SST_x} \\
&= \frac{\sigma^2}{SST_x} \left( \frac{1}{n} \displaystyle\sum\limits_{i=1}^n x_i^2 - (\bar{x})^2 \right) + \frac{\sigma^2 (\bar{x})^2}{SST_x} \\
&=  \frac{\sigma^2 n^{-1} \displaystyle\sum\limits_{i=1}^n x_i^2}{SST_x}
\end{align}
