How to test $ H_0: \mu=0 \quad \text{vs.} \quad H_a: \mu=1$? The Statement of the Problem:

Let $X_1,..., X_n$ be a random sample from $N(\mu, 1)$. Consider testing
  $$ H_0: \mu=0 \quad \text{vs.} \quad H_a: \mu=1. $$
Let the rejection region be $C = \{ \overline X > c\}$
  (a) Find $c$ so that the test has size $\alpha = 0.01$.
  (b) Find the power under $H_a$, i.e. find $\gamma_C(1).$
  (c) Show that $\gamma_c(1) \to  1$ as $n \to \infty$. How do you interpret this? 

Where I Am:
I apologize if this is really basic, but I can't figure out how to do this. I know how to test for alternative hypotheses that are state that the mean is not a value, or larger than or smaller than some value; and I know how to test for the difference between means of different samples, but this doesn't really make sense to me. If someone could shed a little light on this, I'd appreciate it. 
 A: $\bar{X}$, the sample mean, is distributed as $N(0,n^{-1})$ when $H_0$ is
true and as $N(1,n^{-1})$ when $H_1$ is true. (Here, the parameters
are $\mu$ and $\sigma^2$
Thus, we want $c$ to be that number for which $P\{\bar{X} > c \mid H_0\}$
equals $0.01$. But,
$$P\{\bar{X} > c \mid H_0\} =Q\left(c\sqrt{n}\right)$$
where $Q(\cdot)$ is the complementary standard normal distribution function, and a quick look at tables (or resort to an R-gument)
shows that $Q(2.33) \approx 0.01$. Hence,

$$c \approx \frac{2.33}{\sqrt{n}}\tag{1}$$

It follows that 

$$P\{\bar{X} > c \mid H_1\} = Q\left((c-1)\sqrt{n}\right) 
\approx Q\left(2.33 - \sqrt{n}\right)\tag{2}$$

Since the argument of $Q$ in $(2)$ diverges to $-\infty$ as 
$n \to \infty$
(and so the value of $Q(\cdot)$ approaches $1$), we have part (c)
as well. The interpretation is that while the threshold $c$
converges to $0$ as $n\to\infty$ (cf. $(1)$), the variance 
of $\bar{X}$
is converging to $0$, and thus, when $H_1$ is true,
the probability that $\bar{X} \sim N(1,\sqrt{n^{-1}})$ exceeds
the threshold $c$ is close to $1$. Even though that threshold
is quite close to the mean $1$ in absolute terms, it is zillions
of standard deviations away from the mean $1$.
A: Hints:
(a) Under $H_0$ we have $\bar{X} \sim$ normal$(0, 1 / n)$ and so $\bar{X} \sqrt{n} \sim$ normal$(0, 1)$, and so we want to find a $c$ such that $P(\bar{X} \sqrt{n} > c \sqrt{n}) = \alpha$.  It's perhaps a bit confusing that the author framed the question in this way since here $c$ will depend on $n$ in order to get proper size.
(b) You actually can't answer this without knowing $n$, because the power depends on $n$.  I'm beginning to wonder if the author made a mistake.
(c) Suppose we've chosen some $z$ to use as the lower bound of a rejection region of the form $\{ \bar{X} \sqrt{n} > z \}$ (i.e., $z = c \sqrt{n}$).  When $H_1$ is true we can look at the event of rejection in this way $\{ \bar{X} \sqrt{n} > z \} = \{ (\bar{X} - 1)\sqrt{n} + \sqrt{n} > z \} = \{ (\bar{X} - 1)\sqrt{n} > z - \sqrt{n} \}$.  What happens to the probability of this event as $n \to \infty$?
A: For (b), you will  reject $H_0$ if $\frac{\bar{X}-0}{\sigma/\sqrt{n}}\geq Z_{\alpha}$ (where $\sigma=\sqrt{1/n}$ in fact, it should be sample variance)
The power is the probability to reject $H_0$ if the $H_1$ is true
$\gamma_{(c)}(1)=P_{\mu}( \bar{X} \geq Z_{\alpha}\sigma/\sqrt{n} )=P_{\mu}(\frac{\bar{X}-1}{\sigma/ \sqrt{n}} \geq \frac{Z_{\alpha}\sigma/\sqrt{n}-1}{\sigma/\sqrt{n}})=P_{\mu}(\frac{\bar{X}-1}{\sigma/ \sqrt{n}}\geq Z_{\alpha}-\frac{1}{\sigma/\sqrt{n}})=1-\Phi(Z_{\alpha}-n)$ 
($\sigma=\sqrt{1/n},  \Phi $ is CDF of standard normal distribution)
So your power function is $1-\Phi(Z_{\alpha}-n)$. It depends on both $\alpha $ and $n$.
When $\alpha=0.01$
The power is $1-\Phi(1.64485-n)$
You also can see when $n\rightarrow\infty,  \Phi(1.64485-n)\rightarrow0$ so the power will be 1
