When trying to find the normalization constant of $$\int_{-\infty}^{\infty}e^{-x^2}H_m(x)H_n(x) \,dx$$
for the Hermite polynomials $$H_n(x)=(-1)^n e^{x^2}\frac{d^n}{dx^n}e^{-x^2},$$
my reference shows that we can use this equation once in the integral to get $$(-1)^n\int_{-\infty}^{\infty}e^{-x^2}H_m(x) e^{x^2}\frac{d^n}{dx^n}e^{-x^2}\,dx = (-1)^n\int_{-\infty}^{\infty}H_m(x) \frac{d^n}{dx^n}e^{-x^2}\,dx. $$
I'm wondering, why can the exponentials cancel out if $H_m(x)$ is technically an operator (unless they commute)?
$H_m(x)$ is not an operator, or rather, it is an ordinary function of $x$. The $d^n/dx^n$ acts only on the $e^{-x^2}$. So for example $H_1(x) = -e^{x^2} \frac{d}{dx} (e^{-x^2}) = 2x $. There are alternative definitions that do not use derivatives, such as $$ e^{2tx-t^2} = \sum_{n=0}^{\infty} H_n(x) \frac{t^n}{n!}. $$