P.T for any integer n>1, n! Is not a perfect square

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From Bertrand-Chebyshev theorem, if k is a natural there exists at least one prime p such that k<p<2k-2

This prime will divide both (2k-1)! and 2k!.

Now we've got to see to what power it appears which is given by the Legendre function.

$ord_p(n!) = \sum_{k=1}^{\infty} \left[\frac{n}{p^k} \right]$

So if n = 2k or 2k-1 then

$1 \le \frac{n}{p} < \frac{2k}{k} =2$

which means ord_p(n!) =1 .

That means these primes appear to the power 1 in the prime factorisation of n!. But for a perfect square, every prime appears to an even power.

Hence n! cannot be of the form a^b with b>1 i.e it cannot be a perfect power of an integer. In particular it cannot be a perfect square.

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