Let $f(x) = x^n$. Show that $f^{(n)} = n!$ and $f^{(m)} (x) = 0$ for all $m > n$. I'm supposed to use mathematical induction to solve this


*

*Show that $P(1)$ is true 

*Assume $P(K)$ is true

*Show that $P(K+1)$ is true


How do I approach this problem?
 A: You approach the problem by following the steps. Keep in mind that $P(n)$ is the statement 

$f^{(n)} = n!$ and $f^{(m)}(x)=0$ for $m>n$.



*

*Now, if this is $P(n)$, what is $P(1)$?

*How would you prove that $P(1)$ is true?


Once that is done, try to write down $P(k)$ (which is what you assume to be true) and $P(k+1)$ (which is what you want to prove). Write them both down and try to see what the connection between them is. The statements will be very similar.
A: Notice, $n\in\mathbb{Z^+}$:
$$\frac{\text{d}^n}{\text{d}x^n}\left(x^a\right)=\begin{cases}a!\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\text{when}\space n=a\\
0\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\text{when}\space n>a\\
\frac{a!}{(a-n)!}\cdot x^{a-n}\space\space\space\space\space\space\space\space\text{when}\space n<a
\end{cases}$$
Because (example):
$$\frac{\text{d}^3}{\text{d}x^3}\left(x^2\right)=\frac{\text{d}^2}{\text{d}x^2}\left(2x^1\right)=\frac{\text{d}}{\text{d}x}\left(2x^0\right)=0$$
$$\frac{\text{d}^3}{\text{d}x^3}\left(x^3\right)=\frac{\text{d}^2}{\text{d}x^2}\left(3x^2\right)=\frac{\text{d}}{\text{d}x}\left(6x^1\right)=6x^0=6=3!$$
And to proof those, use this:
$$\frac{\text{d}}{\text{d}x}\left(x^a\right)=ax^{a-1}$$
$$\frac{\text{d}^2}{\text{d}x^2}\left(x^a\right)=a(a-1)x^{a-2}$$
$$\frac{\text{d}^3}{\text{d}x^3}\left(x^a\right)=a(a-1)(a-2)x^{a-3}$$
$$\frac{\text{d}^4}{\text{d}x^4}\left(x^a\right)=a(a-1)(a-2)(a-3)x^{a-4}$$
$$\frac{\text{d}^5}{\text{d}x^5}\left(x^a\right)=a(a-1)(a-2)(a-3)(a-4)x^{a-5}$$
