= Solution
Repeated differentiation of the <Gaussian function> gives $D^ne^{-x^2}=P_n(x)e^{-x^2}$, where $P_0=1$ and $P_{n+1}=P_n'-2xP_n$. Induction shows $P_n$ has <degree of a polynomial> $n$ and leading coefficient $(-2)^n$: the differentiated term has lower degree, while multiplication by $-2x$ raises it by one. Thus the <Hermite polynomial> $H_n=(-1)^nP_n$ has \b[degree $n$ and leading coefficient $2^n$].
For $m<n$, apply <integration by parts> $n$ times to the Rodrigues expression:
$$
\int_{-\infty}^{\infty}H_nH_m e^{-x^2}\,dx=(-1)^n\int_{-\infty}^{\infty}H_m D^ne^{-x^2}\,dx=\int_{-\infty}^{\infty}H_m^{(n)}e^{-x^2}\,dx=0.
$$
All boundary terms vanish because each derivative of $e^{-x^2}$ is a <polynomial> times that exponential, and such products tend to zero at either infinity. The same result for $m>n$ follows by symmetry of the <inner product>. The diagonal value is positive; indeed $H_n^{(n)}=2^nn!$ gives
$$
\boxed{(H_n,H_m)=2^nn!\sqrt\pi\,\delta_{nm}.}
$$
This proves the requested <orthogonality> as well as its normalization.
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