Use the moment generating function to obtain the moments of all orders of $X$ if $X \sim N(0, \sigma^2)$
My attempt:
The mgf of a generic $N(\mu, \sigma)^2$ is $M_X(t) = e^{\mu t}e^{(\sigma^2t^2)/2}$. So the mgf of $X$ in this particular case is
$$ M_X(t) = exp(\frac{\sigma^2t^2}{2}), ~~ t \in \mathbb{R} \tag{1} $$
Now you taylor expand the mgf to obtain
$$M_X(t) = \sum_{n = 0}^{\infty}\Biggl( \frac{\sigma^2 t^2}{2} \Biggr)^n\frac{1}{n!} \tag{2} $$
Now the objetive is the simplify the series to the form
$$ \sum_{n = 0}^{\infty}c_n \frac{t^n}{n!} \tag{3} $$
In this form, the expression of $c_n$ will be the expression for $E(X^n)$.
But when simplifying $(2)$, I get:
$$ \frac{\sigma^{2n} t^{2n}}{2^n}\frac{1}{n!} = \frac{\sigma^{2n}(2n)!}{2^n n!} \frac{t^{2n}}{(2n)!} = \frac{\sigma^{n}(n)!}{2^{n/2} (n/2)!} \frac{t^{n}}{(n)!} \tag{4} $$
So you'd conclude that
$$ E(X^n) = \frac{\sigma^{n}(n)!}{2^{n/2} (n/2)!} \tag{5} $$
However, this doesn't agree with the first moment of $X$, which is $0$ since $X \sim N(0, \sigma^2)$.
The argument you transcribed from your teacher is mostly fine, except for a miscommunication in the last equality of $(4)$. Namely, $2n$ was renamed to $n$. It is less confusing if you avoid this renaming. Then, observe that $c_n=0$ if $n$ is odd since there are only even terms in the series. All that is left is to write down a formula for $c_{2n}$, which can be done by using just the first equality in $(4)$.