The only thing I know for this problem is that an integer is a product of primes.
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6First show that $p$ divides $a$ (Euclid's Lemma). – André Nicolas Feb 03 '15 at 17:43
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2Recall that a prime divides a product iff it divides some factor (by The Fundamental Theorem of Arithmetic, i.e. existence and uniqueness of prime factorizations). – Bill Dubuque Feb 03 '15 at 17:45
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1What are the prime factors of $a$? – Joffan Feb 03 '15 at 17:46
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@BillDubuque Yes, then I have p|a, so now I can say p is a prime factor of a, then p^n certainly divides a^n. Thank you guys – SonicFancy Feb 03 '15 at 17:48
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2@SonicFancy Right $,a = pb,$ so $\ a^n = (pb)^n = p^n b^n,$ is divisible by $,p^n.\ $ You can post your answer and accept it. – Bill Dubuque Feb 03 '15 at 17:50
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Thanks for the comments.
Since we have $p$$|$$a^n$ By The Fundamental Theorem of Arithmetic, p|a
then a = pq hence $a^n=$(pq)$^n$ $=$ $p^n$$q^n$.
Now it is obvious that $p^n$$|$$a^n$
SonicFancy
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If $p|a^n$ then the factorization of $a^n$ must look something like this: $a^n = p^n q^n r^n \ldots$ where the various bases on the right hand side are either primes or $1$, and the exponents are all $n$.
Suppose that $p^n$ does NOT divide $a^n$. Then the factorization of $a^n$ must be something like $a^n = p^m q^n r^n \ldots$, with $m < n$. But then the equation would be false, because then $p^m q^n r^n \ldots = p^m b^n$, where $b$ is some integer. If $1 < m < n$, then $p^m b^n$ is an Achilles number (see http://mathworld.wolfram.com/AchillesNumber.html), and not a perfect power, contradicting what we supposed earlier.
This may all seem very abstract, but examples should make it crystal clear. So, for example, we have $p = 2$ and $6^3 = 216$. Since $2$ divides $216$, we know that $2^3$ is also a divisor of $216$.
Of course in my answer so far I've been assuming that we're only dealing with integers. Let's try $p = 2$ like before and $a^3 = 108$. Does $2^3$ also divide $108$? It does not, but $a$ is not an integer. $a \approx 4.7622$ (or $a$ could also be one of two complex numbers that differ only in the sign of the imaginary part).
James47
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