Let p be a prime and let m and n be natural numbers. Prove that p | m(adsbygoogle = window.adsbygoogle || []).push({}); ^{n}implies p^{n}| m^{n}.

Attempt:

Since m^{n}can be written out as a product of primes i.e: p_{1}p_{2}...p_{n}in which p is a divisor.

Raising m^{n}means that there would exist p^{n}primes for each factor of m: m^{n}= m_{1}m_{2}...m_{n}= (p_{1}...p_{n})_{1}(p_{1}...p_{n})_{2}....(p_{1}...p_{n})_{n}= p_{1}^{a1}p_{2}^{a2}...p_{n}^{an}

which means p^{n}| m^{n}.

Is there anything I'm missing to clean it up?

Cheers for the help.

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# Number Theory Question 2

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