On Wikipedia it's said that one could express the Hermite polynomials using the $_1F_1$ function and the following formulae are provided:
$$H_{2n}(x)=(-1)^n \frac{(2n)!}{n!}(_1F_1)(-n,1/2;x^2),$$
$$H_{2n+1}(x)=(-1)^n \frac{(2n+1)!}{n!}2(_1F_1)(-n,3/2;x^2),$$
Even though I've made quite an extensive search online I haven't been able to find some detailed proof of the latter.
If you have some idea on how to prove it, or you know some source where an explanation is given I would be thankful. Thanks in advance!
Using \begin{align} (a)_{-k} &= \frac{(-1)^k}{(1-a)_{k}} \\ (a)_{2 k} &= 2^{2 k} \, \left(\frac{a}{2}\right)_{k} \, \left(\frac{a+1}{2}\right)_{k} \end{align} then, for the even case, \begin{align} H_{2n}(x) &= (2 n)! \, \sum_{k=0}^{n} \frac{(-1)^{n-k} \, (2 x)^{2 k}}{(2 k)! \, (n-k)!} \\ &= \frac{ (-1)^n \, (2 n)!}{n!} \, \sum_{k=0}^{n} \frac{(-1)^{k} \, (2 x)^{2 k}}{(1)_{2 k} \, (n+1)_{-k}} \\ &= \frac{ (-1)^n \, (2 n)!}{n!} \, \sum_{k=0}^{n} \frac{(-n)_{k} \, (2 x)^{2 k}}{k! \, \left(\frac{1}{2}\right)_{k}} \\ &= \frac{ (-1)^n \, (2 n)!}{n!} \, {}_{1}F_{1}\left( -n; \frac{1}{2}; x^2 \right). \end{align}
The case for the odd integer values is obtained in a similar manor.