Prove by induction for$ P(x)$

Question

$P(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$,with $a_n\neq 0$. Using induction on $n$, prove that $P^{(n)}(x) = n!a_n$. So I start with $n=1$ but $P(x) =a_1x$ and $P^{(1)}(x)=1a_1$ are not equal because of the $x$. What do I do with the $x$? Do I have to induce twice?

Answer 1

Note that: \begin{align*} P_0^{(0)}(x) &= P_0(x) = a_0 = 0!a_0 \\ P_1^{(1)}(x) &= P_1'(x) = \frac{d}{dx}[a_1x + a_0] = a_1 = 1!a_1 \\ P_2^{(2)}(x) &= P_2''(x) = \frac{d^2}{dx^2}[a_2x^2 + a_1x + a_0] = 2a_2 = 2!a_2 \\ &~~\vdots \end{align*}