Transition probability between two harmonic oscillator states (Hemite polynomial integration)

Objective:

Show that

$$
int^{infty}_{-infty} x e^{-x^2} H_n(x) H_m(x) dx = pi^{1/2} 2^{n-1} n! delta_{m,n-1} + pi^{1/2} 2^n (n+1)! delta_{m,n+1}
$$

My attempt at this is:

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begin{eqnarray*}
sum^infty_{n=0} frac{2^n s^n u^n}{n!} (s+u) sqrt{pi} &=& sum^infty_{m,n=0}frac{s^m u^n}{m! n!} int x H_m(x) H_n(x) e^{-x^2} dx
sqrt{pi}bigg( sum^infty_{n=0} frac{2^n s^{n+1} u^n}{n!} +sum^infty_{n=0} frac{2^n s^n u^{n+1}}{n!} bigg) &=& sum^infty_{m,n=0}frac{s^m u^n}{m! n!} int x H_m(x) H_n(x) e^{-x^2} dx
end{eqnarray*}

I’m reasonably confident up to here.

begin{eqnarray*}
sqrt{pi}bigg( sum^infty_{n=0} frac{2^n s^{n+1} u^n}{n!} +sum^infty_{n=0} frac{2^n s^n u^{n+1}}{n!} bigg) delta_{m,n+1}&=& delta_{m,n+1}sum^infty_{m,n=0}frac{s^m u^n}{m! n!} int x H_m(x) H_n(x) e^{-x^2} dx
end{eqnarray*}
begin{eqnarray*}
sqrt{pi}bigg( sum^infty_{n=0} frac{2^n s^{n+1} u^n}{n!}big(delta_{m,n+1}frac{m! n!}{s^m u^n}big) +sum^infty_{n=0} frac{2^n s^n u^{n+1}}{n!}big(delta_{m,n+1}frac{m! n!}{s^m u^n}big) bigg) delta_{m,n+1}&=& int x H_m(x) H_n(x) e^{-x^2} dx
end{eqnarray*}

Then I thought maybe I could try re-indexing the n and m in the second $big(delta_{m,n+1}frac{m! n!}{s^m u^n}big)$ term.

begin{eqnarray*}
sqrt{pi}bigg( sum^infty_{n=0} frac{2^n s^{n+1} u^n}{n!}big(frac{(n+1)! n!}{s^{n+1} u^n}big) +sum^infty_{n=0} frac{2^n s^n u^{n+1}}{n!}big(delta_{m,n+1}frac{n! m!}{s^n u^m}big) bigg) delta_{m,n+1}&=& int x H_m(x) H_n(x) e^{-x^2} dx
bigg(sqrt{pi} 2^n (n+1)! + sqrt{pi} sum^infty_{n=0} frac{2^n s^n u^{n+1}}{n!}big(delta_{m,n+1}frac{n! m!}{s^n u^m}big) bigg) delta_{m,n+1}&=& int x H_m(x) H_n(x) e^{-x^2} dx
sqrt{pi} 2^n (n+1)! delta_{m,n+1} + sqrt{pi} sum^infty_{n=0} frac{2^n s^n u^{n+1}}{n!}big(frac{n! (n+1)!}{s^n u^{n+1}}big) delta_{m,n+1}&=& int x H_m(x) H_n(x) e^{-x^2} dx
sqrt{pi} 2^n (n+1)! delta_{m,n+1} + sqrt{pi} 2^n (n+1)! delta_{m,n+1}&=& int x H_m(x) H_n(x) e^{-x^2} dx
sqrt{pi} 2^{n+1} (n+1)! delta_{m,n+1} &=& int x H_m(x) H_n(x) e^{-x^2} dx
end{eqnarray*}

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Obviously the wrong answer. I got the following hint from a friend: "You should take d^n/du^n d^m/ds^m (n-th and m-th derivatives) of the second line, answer has two different Kronecker deltas."

Can anyone specifically help me figure out how to get this $delta_{m,n-1}$ factor? I can’t figure out that one.