Prove there is never a perfect square for N! when N > 2

Here's my attempt at this one......

Let's suppose that N!   = p^2

And  every square can be factored into a product of squared primes

Then....let's suppose that  p is factored as  (a₁)² * (a₂)² * (a₃)² * .....(a_n)²   where each term is a squared prime

And N! = 2 * 3 * 4 *......*N-2 * N-1 * N

So we have that........

2 * 3 * 4 *......*N-2 * N-1 * N  = (a₁)² * (a₂)² * (a₃)² * .....(an)²    

But some of the terms on the left side will not be perfect squares, while every term on the right hand side will be a perfect square

Thus.......the left side cannot be written solely as a product of perfect square primes

Thus.......N!  is not a perfect square

Thank you CPhill

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