= Solution
The coefficient formula from part (b) and the <Leibniz rule> yield <Rodrigues' formula> directly:
$$
p_n(x)=\sum_{j=0}^n\frac{(-1)^j}{j!}\binom njx^j
=\frac{e^x}{n!}\frac{d^n}{dx^n}(e^{-x}x^n).
$$
The leading coefficient is $(-1)^n/n!$, so $p_n^{(n)}=(-1)^n$. Substitute this differentiated expression into one factor in the squared <weighted norm> and apply <integration by parts> $n$ times:
$$
\begin{aligned}
\|p_n\|_w^2
&=\frac1{n!}\int_0^\infty p_n(x)\frac{d^n}{dx^n}(e^{-x}x^n)\,dx\\
&=\frac{(-1)^n}{n!}\int_0^\infty e^{-x}x^n p_n^{(n)}(x)\,dx
=\frac1{n!}\int_0^\infty e^{-x}x^n\,dx
=\boxed{1}.
\end{aligned}
$$
Every endpoint term contains a derivative of $e^{-x}x^n$ of order less than $n$: it vanishes at zero, and its exponential decay beats every <polynomial> at infinity. This proves the <normalization of the ordinary Laguerre polynomials> for all $n$, including $n=0$ without any integrations by parts.
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