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I really need help with proving this using induction.

\begin{align} n! = [n(n-1) &\ldots (n-j+1)](n-j)!\ \ for \ 0 \le j \le n\\ \end{align}

I've tried using brute induction (which is what my prof calls it) and just using n = 1 first, then n = k+1. But I'm not sure what to do with the j? Please help. Anything will be appreciated. Thanks.

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Well...

$(n+1)! = (n+1)*n! = (n+1)[n(n-1).....(n-j+1)](n-j)!; 0 \le j \le n$

$= [(n+1)n(n-1) ...... (n+1 - (j+1) + 1)](n +1 - (j+1))!$

$= [(n+1)n(n-1).........(n + 1 - i+ 1)](n+1 - i)!; 1\le i \le n+1$.

And $(n+1)! = [1](n+1 - 0)!; i=0$.

So .... that's it.

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  • $\begingroup$ This makes a lot of sense! Thanks! $\endgroup$ – b_makio2 Sep 27 '17 at 0:26

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