 Solution

Differentiating the Rodrigues representation directly gives the identityWe prove simultaneously by induction that $H_n^{\prime }=2n{H}_{n-1}$ for $n\ge 1$. It holds for $H_0=1$, $H_1=2x$. Suppose it holds at $n$. Differentiating the preceding identity and using the same identity at $n-1$ givesThe induction is complete. Substituting the derivative identity into the first formula proves the Hermite polynomial three-term recurrence relationThe initial step $H_1=2x{H}_0$ supplies the recurrence at $n=0$ without needing to define $H_{-1}$.