So, by the FTOA, since $a >1$, then a can be broken down into a product of a prime factors, so $a = p_1 \times p_2 \times \dotsm \times p_k$. Then, can I say that since $a$ is multiplied by itself $n$ times, then $a^n$ is $(p_1 \times p_2 \times \dotsm \times p_k)^n$, and since $p|a^n = (p_1 \times p_2 \times \dotsm \times p_k)^n$, then $p|a$ because $p$ divides the product of primes $a$ to the $n$-th power, and therefore, must divide one of the primes in $a$?
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- $\begingroup$ almost - remember that $a$ can be broken down into a product of powers of prime factors, i.e. $a=p_1^i\times p_2^j\times...$ $\endgroup$Mufasa– Mufasa2014-10-19 22:06:27 +00:00Commented Oct 19, 2014 at 22:06
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1 Answer
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Your conclusion is correct. Only to make it more clear you could write $a^n$ as $$a^n=(p_1 \times p_2\times\dotsm\times p_k)^n=p_1^n \times p_2^n\times\dotsm\times p_k^n$$ Now, since $p$ divides $a_n$ and the $p_i$'s are primes, $p$ divides some of the $p_i^n$, that is, there exists at least one $1\le i \le k$ such that $$p|p_i^n$$ but since $p_i$ and $p$ are primes you can conclude that $p|p_i$ and in particular that $p=p_i$. Then it is immediate that $$p|a$$
The reason I wrote "at least one $p_i$" is also what Musafa mentions in his comment. Each prime factor can come up more than once in the factorization.
