# Proving d^l[(x-1)^l]/dx^l = l!

1. Dec 20, 2011

### RK455

1. The problem statement, all variables and given/known data

Sorry I can't do this in latex, the PC I'm on doesn't display it and I can't confidently reel it off without checking so I'll have to do this in an ugly way! :/

Prove:
d^n/dx^n [(x-1)^n] = n!
For integer n

2. Relevant equations

I'm trying induction but it's never been my strong point, I'm not convinced I actually understand how induction works, whenever I try and get my head around it, it seems circular..

3. The attempt at a solution

Testing when n=0 gives (x-1)^0 = 1 = 0! so it holds when n=0

Now assuming true for n=k gives:
d^k/dx^k [(x-1)^k] = k!

If this holds for n=k, that it holds when n=k+1 must be implied by n=k being true:
d^(k+1)/dx^(k+1) [x-1]^(k+1) = d^k/dx^k(d/dx((x-1)^(k+1))) separating the differentiation operator
= d^k/dx^k((k+1)(x-1)^k) evaluating the inner derivative
= (k+1)d^k/dx^k((x-1)^k) bringing the constant (k+1) outside the derivative
= (k+1) k! using the result I'm trying to prove
= (k+1)!

Sorry, that's hideous.. is it OK?

2. Dec 20, 2011

### daveb

Yes, it is hideous, but (unless my eyes glazed over something important), your induction works just fine!

3. Dec 20, 2011

### RK455

OK thanks :)

A further question:
This proof is actually part of my solution to Q20 part C on this paper: