I am given that $X\sim N(0,1)$ and that $Y=e^X$, and I am also given that the kth moment of Y is $E[Y^k]=E[e^{kx}]=M_X(k)=e^{\frac{k^2}{2}}$. This solution suggests that these relationships are self-evident but for the life of me, I can't understand why any of them are true.
Based on my understanding, I would have thought of the kth moment as $\frac{\partial^k}{\partial t^k}$ of $E[e^{tx}]$, that is $\frac{\partial^k}{\partial t^k} \big(e^{tx}\cdot e^x\big)$ but this doesn't seem to make any sense either. I guess I'm struggling with the very concept of the kth moment...
Yes, self evident: I explain.
If $X\sim N(\mu;\sigma^2)$
$\mathbb{E}[Y^t]=\mathbb{E}[e^{Xt}]=e^{t\mu+\frac {t^2\sigma^2}{2}}$
This result has been obtained using the definition of Moment Generating Function of the Gaussian distribution.
I your Gaussian is standard, the result is
$$\mu_t(Y)=e^{\frac {t^2}{2}}$$
Note also that $Y=e^X$ follows a lognormal distribution
EDIT: this edit is to answer to the latest comment of the O.P.
Although it can be calculate without difficulty, you can trust that the MGF of a Standard Gaussian is the following function
$M_X(t)=e^{\frac{t^2}{2}}$
This function, as per definition, is the following expected value
$\mathbb{E}[e^{Xt}]$
[No difference if we say $M_X(k)=e^{\frac{k^2}{2}}$]
Just observe that
$e^{Xt}=[e^{X}]^t$
thus, setting the new rv $Y=e^X$ it is self-evident that
$\mathbb{E}[e^{Xt}]=\mathbb{E}[Y^t]$
In other words: the moments of the Log Normal are given by the MGF of the gaussian
EXAMPLE
$\mathbb{E}[Y]=e^{\frac{1}{2}}$
$\mathbb{V}[Y]=e^2-[e^{\frac{1}{2}}]^2=e(e-1)$
You can double check these result here