How should I prove $\int_0^\infty\frac{d^j}{dx^j}(x^je^{-x})dx=0$?

Question

Context: Using the weighted inner product definition $$\langle f,g\rangle_{w(x)}=\int_a^bf(x)g(x)w(x)dx$$ for real valued functions $f(x),g(x),w(x)$, I wish to show that the following two functions are orthogonal for the interval $[0,\infty)$ with weight $w(x)=e^{-x}$: $f(x) = 1$, $g(x) = e^x \frac{d^j}{dx^j}(x^je^{-x})$. That is, I wish to show that the weighted inner product yields $0$.

More context: Here, $g(x)$ is a Laguerre polynomial $L_j(x)$ (and $f(x)$ is also a Laguerre polynomial, namely the first one $L_0(x)$.) Laguerre polynomials are orthogonal for the given above weight function on the above interval, and I am trying to show this in cases, first for $L_0, L_{j}, j>0$ and then I will try and move on to the case of $L_k, L_j, 0<k<j$.

Attempt: I have shown that $\frac{d^j}{dx^j}(x^je^{-x})=x^0q_j(x)e^{-x}$ for all $j = 0, 1,\dots$ where $q_j(x)$ is a degree $j$ polynomial. This yields that I need to show that $$\int_0^\infty e^{-x}q_j(x)dx$$ is zero, but for arbitrary degree $j$ polynomial $q_j$ this does not seem to be true.

Answer 1

By the Leibniz Theorem, $$\frac{d^{j}}{dx^{j}}[x^{j}e^{-x}] = \sum_{i=0}^{j}\binom{j}{i}(-1)^{j-i}e^{-x}\frac{j!}{(j-i)!}x^{j-i}$$ And so, $$\int_{0}^{\infty}\frac{d^{j}}{dx^{j}}[x^{j}e^{-x}] \ dx = \int_{0}^{\infty}\sum_{i=0}^{j}\binom{j}{i}(-1)^{j-i}e^{-x}\frac{j!}{(j-i)!}x^{j-i} \ dx$$ $$= j!\sum_{i=0}^{j}\binom{j}{i}(-1)^{j-1}\frac{1}{(j-i)!}\int_{0}^{\infty}e^{-x}x^{j-i} \ dx$$ $$ = j!\sum_{i=0}^{j}\binom{j}{i}(-1)^{j-1}\frac{1}{(j-i)!}\Gamma(j-i+1)$$ $$ = j!\sum_{i=0}^{j}\binom{j}{i}(-1)^{j-1}\frac{1}{(j-i)!}(j-i)!$$ $$ = j!\sum_{i=0}^{j}\binom{j}{i}(-1)^{j-1}$$ $$ = -j!(1 - 1)^{j}$$ $$ = 0$$