= Solution
By the <Leibniz rule>,
$$
\frac{d^n}{dx^n}\left(x^{n+1/2}e^{-x}\right)
=e^{-x}\sum_{j=0}^n\binom nj(-1)^{n-j}
\frac{d^j}{dx^j}x^{n+1/2}.
$$
After multiplication by $(-1)^nx^{-1/2}e^x$, the term $j=0$ is $x^n$, and every other term has lower degree. Thus $p_n$ is a <monic polynomial> of degree $n$. Direct calculation gives
$$
\boxed{
p_0(x)=1,\qquad
p_1(x)=x-\frac32,\qquad
p_2(x)=x^2-5x+\frac{15}{4}
}.
$$
These are the monic normalization of the <Generalized Laguerre polynomial>[generalized Laguerre polynomials] with parameter $1/2$.
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