Let $X\sim N_p(\mu,\Sigma)$, where $N_p$ denotes the mulvariate normal distribution of dimension $p$ and let $\alpha\in\mathbb{R}^p$. Prove that: $$ E[(\alpha'(X-\mu))^k] = \left\{\begin{array}{lc} \frac{(2m)!}{2^mm!}(\alpha'\Sigma\alpha)^m, & \text{ si } k=2m\\ 0, & \text{ si } k=2m-1 \end{array}\right. $$ where $\alpha'$ is the transpose of $\alpha$.
In order to prove this, I have found that $X\sim N_p(0,\Sigma)$, then $$ E[X^k] = \left\{\begin{array}{lc} \sigma^k(k-1)!!, & \text{ si } k \text{ even}\\ 0, & \text{ si } k \text{ odd} \end{array}\right. $$ where $!!$ denotes the double factorial, defined by the product of all integers between 1 and $n$ with the same parity.
I think that result should help a lot with the proof, but I do not know where to use it. I mean, I have tried to expand the term $E[(\alpha'(X-\mu))^k]$ to find, at some point, the term $E[X^k]$ and use that result.
Any suggestion?
Hint: show that $\alpha'(X-\mu) \sim N(0, \alpha' \Sigma \alpha)$. Then with $\sigma^2 := \alpha' \Sigma \alpha$, the result you found regarding moments of univariate normal random variables implies $$E[(\alpha'(X-\mu))^k] = \begin{cases}(\alpha' \Sigma \alpha)^{k/2} (k-1)!! & \text{$k$ even} \\ 0 & \text{$k$ odd}\end{cases}$$ which the desired result.