= Solution
Differentiating the Rodrigues representation directly gives the identity
$$
H_{n+1}=2xH_n-H_n'.
$$
We prove simultaneously by <induction> that $H_n'=2nH_{n-1}$ for $n\geq1$. It holds for $H_0=1$, $H_1=2x$. Suppose it holds at $n$. Differentiating the preceding identity and using the same identity at $n-1$ gives
$$
\begin{aligned}
H_{n+1}'&=2H_n+2xH_n'-H_n''\\
&=2H_n+4nxH_{n-1}-2nH_{n-1}'\\
&=2H_n+4nxH_{n-1}-2n(2xH_{n-1}-H_n)=2(n+1)H_n.
\end{aligned}
$$
The induction is complete. Substituting the derivative identity into the first formula proves the <Hermite polynomial> three-term <recurrence relation>
$$
\boxed{H_{n+1}(x)=2xH_n(x)-2nH_{n-1}(x)\quad(n\geq1).}
$$
The initial step $H_1=2xH_0$ supplies the recurrence at $n=0$ without needing to define $H_{-1}$.
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