Induction differential proof So I was doing an induction proof that contained a differential. Now I got through most of it but lastly in order to complete the proof I needed to prove 
$$\frac{d^k  }{dx^k}x^k=k!$$
Now I don't need to do this via induction or anything it's just that inside the actual induction proof this statement was there and in order to prove the $(k+1)$-th term I need to show this. 
I can see how this would happen:
$\frac{d  }{dx}x^k = k(x^{k-1})$
$\frac{d^2  }{dx^2}x^k  = k(k-1)x^{k-2}$
..
...
..
Until eventually I would get
$k\cdot (k-1)\cdot (k-2)\cdot (k-3)\cdot ... \cdot x^{k-k}$
Which in turn would be $k!$
However is there any better way of showing this, like a more formal proof or better strucutred one ? 
 A: Another way : suppose $\frac{d^n}{dx^n}(x^n) = n!$ for some $n\in\mathbf N$. Then :
$$\frac{d^{n+1}}{dx^{n+1}}(x^{n+1}) = \frac{d^n}{dx^n}\left(\frac{d}{dx}(x^{n+1})\right) = \frac{d^n}{dx^n}((n+1)x^n) = (n+1)\frac{d^n}{dx^n}(x^n) = (n+1)n! = (n+1)!$$
There you have to know the first derivative of $x^n$ for any $n$, which can be proved... by induction, knowing the derivative of a product and the derivative of $x$ and $1$ :-)
A: You wrote that you don't need to do this via induction, but it's a good way of doing it. What I mean by this is that you can prove by induction on $k$ that $\frac{\mathrm d^k}{\mathrm dx^k}x^n=n(n-1)\cdots(n-k+1)x^{n-k}$, where $n\geqslant k$. If $k=1$, this is just the assertion that $\frac{\mathrm d}{\mathrm dx}x=1$. Suppose that it is true for some $k$. Then\begin{align}\frac{\mathrm d^{k+1}}{\mathrm dx^{k+1}}x^n&=\frac{\mathrm d}{\mathrm dx}\left(\frac{\mathrm d^k}{\mathrm dx^k}x^n\right)\\&=\frac{\mathrm d}{\mathrm dx}\left(n(n-1)\cdots(n-k+1)x^{n-k}\right)\\&=n(n-1)\cdots(n-k+1)(n-k)x^{n-k-1}\end{align}
