Why are odd moments of Normal distribution equal to 0

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I am studying the topic of MGF and have come across the result that odd moments of $N(0,1)$ are $0,$ yet I cannot see why.

Given $\mu_k$ the $k$-th moment and from the fact that:

\begin{equation} M(t)=e^{t^2/2} = \sum_{k=0}^{\infty}\frac{t^{2k}}{2^k k!} \end{equation} They claim that

Odd powers of $t$ are all $0$

Therefore $\mu_k = 0$ and $\mu_{2k} = \frac{(2k)!}{2^k k!}$

I cannot see why that is true.

Thanks a lot for any hint.