Calculate $\mathbb{E}[M_{\alpha}^{p}(t)]$ for all $p>0$ and $t>0$, where $M_{\alpha}(t):=e^{\alpha W_t-\frac{\alpha^2}{2}t}$, $t\ge 0$ I am going through this solved problem but I don't understand some steps. My professor is notorious for making errors very often so don't hold back if you think he's wrong... Or if I am wrong. I am going to write my questions in bold to make it easier. Thanks a lot in advance for any help you can provide!
EXERCISE: Let us consider the following martingale
$$M_{\alpha}(t):=e^{\alpha W_t-\frac{\alpha^2}{2}t},$$
where $t\ge 0$, $\alpha\ge 0$.
A) Check that $\mathbb{E}M_{\alpha}(t)=1$.
B) Calculate $\mathbb{E}[M_{\alpha}^{p}(t)]$ for all $p>0$ and $t>0$.
ANSWER:
A) We observe that $M_{\alpha}(t)$ is a martingale. Hence,
$$\mathbb{E}M_{\alpha}(t)=\mathbb{E}M_{\alpha}(0)=1$$
First Question: Why? This is what I get instead:
$$\mathbb{E}M_{\alpha}(t)=\mathbb{E}\left(e^{\alpha W_t-\frac{\alpha^2}{2}t}\right)=\mathbb{E}\left(\frac{e^{\alpha W_t}}{e^{\frac{\alpha^2}{2}t}}\right)=\frac{e^0}{e^{\frac{\alpha^2}{2}t}}=\frac{1}{e^{\frac{\alpha^2}{2}t}}$$
$$\mathbb{E}M_{\alpha}(0)=\frac{e^0}{e^0}=1$$
B)
It follows that
$$\mathbb{E}M_{\alpha}^{p}(t)=\mathbb{E}\left[M_{p\alpha}(t)e^{\frac{\alpha^2 p^2}{2}t-\frac{\alpha^2p}{2}t}\right]=e^{\frac{\alpha^2}{2}(p^2-p)t}$$
Second question: I don't understand the first equality above. In particular, why is this the result in the second equality below?
$$\mathbb{E}M_{\alpha}^{p}(t)=\mathbb{E}\left(e^{\alpha W_t-\frac{\alpha^2}{2}t}\right)^p=\mathbb{E}\left[M_{p\alpha}(t)e^{\frac{\alpha^2 p^2}{2}t-\frac{\alpha^2p}{2}t}\right]=e^{\frac{\alpha^2}{2}(p^2-p)t}$$
 A: For reference what you have is an exponential martingale.
For A) I would just go about it that way
$$ E[M_\alpha(t)]= e^{-\alpha^2/t}E[e^{\alpha W_t}]=e^{-\alpha^2/t}\int_Re^{\alpha \sqrt tZ} \phi(z) dz$$ where $\phi(.)$ is the normal pdf, compute the integral, done. B) is a generalization of the preceding steps.
Second question: $$E[M_\alpha(t)^p]= e^{-\alpha^2pt/2}E[e^{p\alpha\sqrt t Z}]=e^{-\alpha^2pt/2}e^{p^2\alpha^2t/2}=e^{\alpha^2t/2(p^2-p)}$$
 the second equality is motivated by the definition of the MGF for a std normal variable
A: You can also approach this problem using Ito's formula. Let $X = W_t$, $g(x)=e^{\alpha px}$, then $Y_{\alpha}(t) = g(X_t) = e^{\alpha p W_t}$. Now the Ito's formula gives
$$
dY_{\alpha}(t)=\alpha pY_{\alpha}(t)dW_t+\frac{\alpha^2p^2}{2}Y_{\alpha}(t)dt
$$
so integration gives
$$
Y_{\alpha}(t)-Y_{\alpha}(0)=\alpha p\int_0^tY_{\alpha}(\tau)dW_{\tau}+\frac{\alpha^2p^2}{2}\int_0^tY_{\alpha}(\tau)\,d\tau
$$
Taking expectation we have
$$
E[Y_{\alpha}(t)]-E[Y_{\alpha}(0)] = \frac{\alpha^2p^2}{2}\int_0^tE[Y_{\alpha}(\tau)]\,d\tau
$$
by noticing that the term
$$
E[\int_0^tY_{\alpha}(\tau)dW_{\tau}]=0
$$
because $Y_{\alpha}(\tau)\in\mathcal{F}_{\tau}$. This is a deterministic ODE, so one can solve it and get
$$
E[Y_{\alpha}(t)]=e^{\frac{\alpha^2p^2}{2} t}
$$ 
So now we can conclude that
$$
E[M^p_{\alpha}(t)]=E[Y_{\alpha}(t)]\cdot e^{-\frac{\alpha^2p}{2}t}=e^{\frac{\alpha^2}{2}(p^2-p)t}
$$
