show (n^2)! is divisible by (n!)^n+1(adsbygoogle = window.adsbygoogle || []).push({});

base case 1 divides 1

so I did the induction and got

(n^2+2n)!=((n+1)^n * (n!) *(n!)^n+1

(n^2+2n)*.......*(n^2)!=((n+1)^n * (n!) *(n!)^n+1

can we conclude that (n^2)! is divisible by (n!)^n+1 for all n from there because we have the bold terms on both sides?

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# Homework Help: Induction Division

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