I don't understand the variance of the binomial I feel really dumb even asking such a basic question but here goes:
If I have a random variable $X$ that can take values $0$ and $1$, with $P(X=1) = p$ and $P(X=0) = 1-p$, then if I draw $n$ samples out of it, I'll get a binomial distribution.
The mean of the distribution is
$\mu = np = E(X)$
The variance of the distribution is
$\sigma^2 = np(1-p)$
Here is where my trouble begins:
Variance is defined by $\sigma^2 = E(X^2) - E(X)^2$. Because the square of the two possible $X$ outcomes don't change anything ($0^2 = 0$ and $1^2 = 1$), that means $E(X^2) = E(X)$, so that means
$\sigma^2 = E(X^2) - E(X)^2 = E(X) - E(X)^2 = np - n^2p^2 = np(1-np) \neq np(1-p)$
Where does the extra $n$ go? As you can probably tell I am not very good at stats so please don't use complicated terminology :s
 A: A random variable $X$ taking values $0$ and $1$ with probabilities $P(X=1)=p$ and $P(X=0)=1-p$ is called a Bernoulli random variable with parameter $p$. This random variable has
\begin{eqnarray*}
E(X)&=&0\cdot (1-p) + 1\cdot p = p\\
E(X^2)&=&0^2\cdot(1-p) + 1^2\cdot p = p\\
Var(X)&=& E(X^2)-(E(X))^2=p-p^2=p(1-p)
\end{eqnarray*}
Suppose you have a random sample $X_{1},X_{2},\cdots,X_{n}$ of size $n$ from $Bernoulli(p)$, and define a new random variable $Y=X_{1}+X_{2}+\cdots +X_{n}$, then the distribution of $Y$ is called Binomial, whose parameters are $n$ and $p$.
The mean and variance of the Binomial random variable Y is given by
\begin{eqnarray*}
E(Y)&=&E(X_{1}+X_{2}+\cdots + X_{n})=\underbrace{ p+p+\cdots +p}_{n}=np\\
Var(Y)&=& Var(X_{1}+X_{2}+\cdots + X_{n})=Var(X_{1})+Var(X_{2})+\cdots + Var(X_{n})\\
& &\text{ (as $X_{i}$'s are independent)} \\
&=&\underbrace{p(1-p)+p(1-p)+\cdots+ p(1-p)}_{n}\quad
\text{ (as $X_{i}$'s are identically distributed)} \\
&=&np(1-p)
\end{eqnarray*} 
A: Two mistakes in your proving process:
1: $X$ in first paragraph has different definition comparing with $X$ in the rest of article.
2: Under the condition that $X$ ~ $Bin(p, n)$, $E(X^2) \ne E(X)$. Try to work from $E(X^2) = \sum {\left(x^2\Pr(X=x)\right)}$
