= Solution
Write $\widetilde P_j(x)=x^{j-1}+\sum_{r=0}^{j-2}b_{rj}x^r$. The evaluation <matrix> $[\widetilde P_j(x_i)]$ is $V_m B$, where $B$ is upper triangular with diagonal entries $1$. Its <determinant> is therefore unchanged. This is <Vandermonde determinant invariance under monic basis change>:
$$
\det[\widetilde P_j(x_i)]_{i,j=1}^m=\prod_{k<l}(x_l-x_k).
$$
For $w(x)=x^a e^{-x}$ on $[0,\infty)$, multiply row $i$ by $\sqrt{w(x_i)}$. The requested weighted product becomes the <weighted Vandermonde determinant> square
$$
\boxed{\prod_{j<k}(x_j-x_k)^2\prod_{i=1}^m w(x_i)=\left(\det[\sqrt{w(x_i)}\,\widetilde P_j(x_i)]_{i,j=1}^m\right)^2.}
$$
The sign difference between the two orientations of the Vandermonde product disappears on squaring. For $a=0$ interpret the weight at zero by continuity, so $w(0)=1$; for $a>0$, $w(0)=0$.
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