Suppose a continuous random variable has odd moments = zero and even moments as follows
$$E[X^{2n}] = \frac{(2n)!}{2^n n!}$$
Then the mgf is, by Maclaurin series expansion where we can switch sum and integral by nonnegativity of integrand,
$$E[e^{tX}] = \sum_{n = 1}^{\infty} \frac{1}{(2n)!} \frac{(2n)! t^{2n}}{2^n n!}$$
whose closed form is standard normal as pointed out in answer, sort of.
You made an error. You should be careful when writing sums.
This is the central normal distribution you have to guess.
Indeed let $G$ be a gaussian $G\sim\mathcal{N}(0,1)$:
$$\mathbb{E}[e^{tG}]=e^{\frac{t^2}{2}}=\sum_{n\geq 0}t^n\frac{\mathbb{E}[G^n]}{n!}=\sum_{k\geq 0}\frac{t^{2k}}{k!2^k}$$ I identify same powers of $t^k$ If $n=2k+1$, then by rhs I have $E[G^{2k+1}]=0$ if $n=2k$, then I got: $$\frac{E[G^n]}{n!}=\frac{E[G^{2k}]}{(2k)!}=\frac{1}{k!2^k}$$
https://fr.wikipedia.org/wiki/Loi_normale#Moments