proof by induction (derivative of $x^n=n!$)

Question

I'm trying to use induction to prove that the $n$th derivative of $x^{n}$ is $n!$. So, $$\frac{d^{n}}{dx^{n}} x^{n}=n!$$

This is what I've done so far,

Base Case: $n=1 \implies$ $$\frac{d^{1}}{dx^{1}} x^{1}=1!$$ So, $1=1$.

Assume: $n=k \implies$ $$\frac{d^{k}}{dx^{k}} x^{k}=k!$$

End Goal (what we want to show): $$\frac{d^{k+1}}{dx^{k+1}} x^{k+1}=(k+1)!$$

So, $$\frac{d^{k}}{dx^{k}} x^{k}+x^{k+1}$$ $$= kx^{k-1}+x^{k+1}$$

But I'm not sure where to go from here or if this was the right way to approach this.

Answer 1

Your statement may be: $\frac{d^{n}}{dx^{n}} x^{n}=n!$ for $n\in\mathbb{N}.$

The last step of your induction should be:

$$\frac{d^{k+1}}{dx^{k+1}} x^{k+1}=\frac{d^{k}}{dx^{k}}(\frac{d}{dx} x^{k+1})=\frac{d^{k}}{dx^{k}}((k+1)x^k)=(k+1)\cdot k! = (k+1)!.$$

Answer 2

You are using the wrong rule. The $k+1$ exponential derivative is decomposed as: $$\begin{align}\dfrac{\mathrm d^{k+1}~x^{k+1}}{\mathrm d ~x~^{k+1}~~~~} ~&=~ \dfrac{\mathrm d^k ~~~~}{\mathrm d~x~^k}\left(\dfrac{\mathrm d~x^{k+1}}{\mathrm d~x~~~~~~}\right)\\[1ex] ~&=~\dfrac{\mathrm d^k \left((k+1)~x^{k}\right)}{\mathrm d~x~^k\qquad\qquad} \\[1ex] & = ~(k+1)\cdot\dfrac{\mathrm d^k~x^k}{\mathrm d~x~^k }\end{align}$$

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Sometimes I think it would be helpful to write the $n$th derivative as $\dfrac{\mathrm d^n~~~~}{(\mathrm d ~x)^n}$ just to be clear that we are deriving $n$-times with respect to $x$, rather than deriving once with respect to $x^n$.