Use of a substitution to prove that $e^{2xt-t^2}$ is the exponential generating function of the Hermite polynomials

If $u=x-t$ then $t=x-u$ and $dt=d\left(x-u\right)=dx-du=-du$ ($x$ here is like constant).
Also $d^{n}t=\left(-1\right)^{n}d^{n}u$ because each differentiation changes sign to opposite.

$f^{\left(n\right)}\left(t\right)=\left[\frac{d^{n}}{dt^{n}}e^{-\left(x-t\right)^{2}}\right]_{t=0}=\left[\frac{d^{n}}{\left(-1\right)^{n}d^{n}u}e^{-u^{2}}\right]_{x-u=0}=\left[\left(-1\right)^{n}\frac{d^{n}}{d^{n}u}e^{-u^{2}}\right]_{u=x}=\\=\left(-1\right)^{n}\frac{d^{n}}{d^{n}x}e^{-x^{2}}=e^{-x^{2}}\cdot\left(-1\right)^{n}e^{-x^{2}}\frac{d^{n}}{d^{n}x}e^{-x^{2}}=e^{-x^{2}}H_{n}\left(x\right)$

Last part I took from JamesT (see comment above).