Let $X_1,X_2,...$ be i.i.d. RVs and let N be another independent RV which takes values from nonnegative integers. Let $Y=\sum_{k=1}^N X_k$ Using the fact that $M_Y(t)=M_N(logM_{X_1}(t))$, derive $$E[Y]=E[N]E[X_1]$$

I don't quite see how to use the hint to prove this. Any ideas?
Differentiate both sides in $t$ and set $t=0$ to get your answer

$\frac{d}{dt}M_Y(t)|_{t=0} = E[Y]$.

$\frac{d}{dt}M_N(\log M_{X_1}(t))|_{t=0} = \frac{M_{X_1}'(t)}{M_{X_1}(t)}M_N'(\log M_{X_1}(t))|_{t=0} = \frac{E[X]}{1}M_N'(0) = E[X]E[N]$
There is a simpler way to do this BTW, just for the record. 

$$E(Y) = E(E(Y|N))$$

$$E(Y|N) = E(\sum_{k=1}^{N} X_{k})$$

$$=\sum_{k=1}^{N} E(X_{k})$$
$$=N\times E(X_1)$$

Therefore, 

$$E(E(Y|N)) = E(X_1)\times E(N)$$