Divisibility of (n^2)! by (n!)^n+1: Proven Through Induction

[Punkyc7]

show (n^2)! is divisible by (n!)^n+1

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?

 

[SammyS]

The inductive step:

Assume that for some k, (k²)! is divisible by (k!)^k+1.

Now prove that ((k+1)²)! is divisible by ((k+1)!)^k+2.

 

[Punkyc7]

I have done all the work its just a lot to type out so I put the last line down. I haven't taken number theory and I am wondering if the point I got it to can I concluded that the it is divisible because of our induction hypothesis