generating function of Hermite polynomials

We start from the definition of Hermite polynomials via their http://planetmath.org/node/11983Rodrigues formula

$H_n(z):={(-1)}^n{e}^{z^2}{\displaystyle \frac{d^n}{d{z}^n}}{e}^{-z^2}\mathit{\hspace{1em}\hspace{1em}}(n=\mathrm{\hspace{0.33em}0},\mathrm{\hspace{0.17em}1},\mathrm{\hspace{0.17em}2},\mathrm{\dots }).$(1)

The consequence

$f^{(n)}(z)={\displaystyle \frac{n!}{2\pi i}}{\displaystyle {\oint }_C}{\displaystyle \frac{f(\zeta )}{{(\zeta -z)}^{n+1}}}𝑑\zeta$(2)

of http://planetmath.org/node/1150Cauchy integral formula allows to write (1) as the complex integral

$$H_n(z)={(-1)}^n\frac{n!}{2i\pi }{\oint }_C\frac{e^{z^2-{\zeta }^2}}{{(\zeta -z)}^{n+1}}𝑑\zeta ,$$

where $C$ is any contour around the point $z$ and the direction is anticlockwise.  The http://planetmath.org/node/11373substitution  $z-\zeta :=t$  here yields

$$H_n(z)=\frac{n!}{2i\pi }{\oint }_{C^{\prime }}\frac{e^{z^2-{(z-t)}^2}}{t^{n+1}}𝑑t,$$

where the contour $C^{\prime }$ goes round the origin.  Accordingly, by (2) we can infer that

$$H_n(z)={\left[\frac{d^n}{d{t}^n}{e}^{z^2-{(z-t)}^2}\right]}_{t=0},$$

whence we have found the generating function

$$e^{z^2-{(z-t)}^2}=\sum _{n=0}^{\mathrm{\infty }}{H}_n(z)\frac{t^n}{n!}$$

of the Hermite polynomials.