= Solution
Multiplication by $x$ transforms to $i\,d/d\lambda$. Write $p_n(x)=\sum_{j=0}^na_jx^j$ and use $\widehat g=cg$ for the Gaussian $g=e^{-x^2/2}$. Then
$$
\widehat{f_n}(\lambda)=c\sum_{j=0}^na_ji^j\frac{d^j}{d\lambda^j}e^{-\lambda^2/2}
=q_n(\lambda)e^{-\lambda^2/2}.
$$
Successive differentiation of the Gaussian gives a <polynomial> of degree $j$ with leading term $(-1)^j\lambda^j$ multiplying that same Gaussian. Thus $q_n$ has leading <coefficient> $ca_n(-i)^n\ne0$. This proves \b[$q_n$ is a <polynomial> of degree exactly $n$], rather than merely at most $n$.
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