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I need to prove that $d^n(x^n)/dx^n = n!$ by induction.

Any help?

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I think you want to write $d^n(x^n)/dx^n$ – Amr Jun 25 '13 at 11:59

Hint: Are you familiar with proofs by induction? Well, the induction step could be written as $$d^{n+1}(x^{n+1}) / dx^{n+1} = d^n \left(\frac{d(x^{n+1})} {dx}\right) /dx^n $$

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Yes i am, but this is actually the step was having trouble with. – PooperScooper Jun 25 '13 at 12:10
Well, $d(x^{n+1}) / dx = (n+1) x^n$, so you have exactly what you need for the induction to work, right? – Pedro M. Jun 25 '13 at 12:14

Let's start the induction at $n=1$ (could also be done at $n=0$ but that might be a bit confusing). $$\dfrac{d^1}{dx^1}x^1 = \dfrac{d}{dx}x = 1 = 1! $$ Now let's try at any $n>1$: $$\dfrac{d^n}{dx^n}x^n = \dfrac{d^{n-1}}{dx^{n-1}} \lbrace \dfrac{d}{dx}x^n \rbrace = \dfrac{d^{n-1}}{dx^{n-1}} \lbrace nx^{n-1} \rbrace $$ move the constant out of the bracket, and then use the induction step (for $n-1$): $$ =n(n-1)!=n! $$

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If $f_{n} = x^{n}$, noting that $f_{0} = 1$. Taking an initial step to show that $\dfrac{d^1}{dx^1}f_{n} = nf_{n-1}$ and continuing with a second: $\dfrac{d^2}{dx^2}f_{n} = n(n-1)f_{n-2}$. Now taking the nth step you can see that $\dfrac{d^n}{dx^n}f_{n} = n(n-1)(n-2)(n-3)...(n-(n-1))f_{0}$

Inductive step

if $k = n + 1$ then $\dfrac{d^{k}}{dx^{k}}f_{k} = (n+1)n\dfrac{d^{k-1}}{dx^{k-1}}f_{n} = k!$

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