Is This a Sufficient Proof for Perfect Squares with Even Exponents?

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n=p1r1...pkrk

In order for p to be a perfect square, r must be even. Therefore

n=p12h1...pk2hk

taking the square root of both sides I'm just left with

n=p1h1...pkhk

Does this work as a proof that n is a perfect square if r is even? It's a homework problem and I'm not sure if this is sufficient or correct at all.
 
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If the p's are distinct primes, it's true the r's must be even for n to be a perfect square, but you haven't proved it. You haven't even stated the problem coherently. What's p?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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