Prove that $P^{(n)}(x) = a_{n}n!$ if $P(x) = a_{n}\cdot x^{n} + a_{n-1}\cdot x^{n-1} + \cdots + a_{0}$ Let $P(x) = a_{n}\cdot x^{n} + a_{n-1}\cdot x^{n-1} + \cdots + a_{0}$ be a polynomial with real coefficients. How can we prove that its $n$th derivative is:$$P^{(n)}(x) = a_{n}n!$$
 A: Hint:
From here observe that $$\frac{dx^m}{d^nx}=\frac{m!}{(m-n)!}x^{m-n}$$ (Proof)
$$\text{ So, }\frac{dx^m}{d^nx}=0 \text{ if }0\le m<n\text { and } \frac{dx^n}{d^nx}=\frac{n!}{(n-n)!}x^{n-n}=n!$$
A: By induction: you have $(a_1 x+a_0)' = a_1\cdot 1!$. Thus, the claim holds for $n = 1$. Assume now that the claim holds for $n-1$ and let us show that it implies the claim for $n$.
Let 
$$
  P_n(x) = \sum_{k=0}^n a_kx^k
$$
be any polynomial of the degree $n$, then
$$
  P_n'(x) = \sum_{k=1}^{n}k\cdot a_{k}x^{k-1} = \sum_{k=0}^{n-1}b_k x^k =: P_{n-1}(x)
$$
is a polynomial of the degree $n-1$ with the main term $b_{n-1} = na_{n}$. We know that
$$
  P^{(n)}_n = P^{(n-1)}_{n-1} = (n-1)!b_{n-1} = n! a_n
$$
A: Let you polynomial be $$p(x)=\sum_{k=0}^n a_kx^k$$
Note that for any polynomial $q$ of degree $m<n$, we have that $$q^{(m)}\equiv 0$$ (i.e. it is the zero polynomial)
Moreover, note that $$p(x)=a_nx^n+q(x)$$ where $\operatorname{deg}(q)\leq n-1$. Then $$p^{(n)}(x)=a_n(x^n)^{(n)}+0=a_n(x^n)^{(n)}$$ 
So, you can prove by induction two things now
$1.$ If $q$ is a polynomial with degree $m<n$, $q^{(n)}=0$.
$2.$ If $f(x)=x^n$ then $f^{(n)}=n!$ 
and you'll be done.
A: The derivative is linear and lower monomials vanish so it suffices to prove $\left(\frac{d}{dx}\right)^n x^n = n!$, we prove a slightly stronger statement by induction:
Theorem $\left(\frac{d}{dx}\right)^k x^n = \frac{n!}{(n-k)!} x^{n-k}$.
proof: $k=0$ is trivial, suppose it's true for $k$ we show it for $k+1$ by differentating RHS once $$\frac{d}{dx} \frac{n!}{(n-k)!} x^{n-k} = \frac{n!}{(n-k-1)!}x^{n-k-1}.$$
