I found out that $p^n$ only has the factors ${p^{n-1}, p^{n-2}, \ldots p^0=1}$, is there a reason why?

I don't understand all the sarcastic answers. I think this is a great question.

The big fact about prime factorisation (which DonAntonio is referring to in code) is not that it exists, but that it's unique - there is only one way to factorise a number into primes. That is, no matter how you factorise a number, you'll always get the same prime factors (and the same number of each).

So whenever you factorise 252, you'll get two 2s, two 3s and a 7, and nothing else. So, is 6 a factor of 252? Yes - 6 factorises into one 2 and one 3, and obviously 2*3 divides 2*2*3*3*7. Is 8 a factor of 252? No - 8 is 2*2*2, but we don't have 'enough' factors of 2 in 252. Is 11 a factor of 252? No - we don't have enough (or any!) factors of 11 in 252. The "uniqueness" ensures that, in order for m to divide n, we can simply factorise both into primes and check whether n contains all the primes that m does (and maybe more) or not.

So, back to your question. Does 6 divide $5^{100}$? No - the right hand side factorises into $5\times 5\times \dots \times 5$, which doesn't contain enough factors of 2 (or 3) for 6 to divide it. And the same is true more generally too.


Let $ab=p^n$. Consider the prime factorization of the two terms on the left hand side. If any prime other than $p$ appears on the left, say $q$, then it appears as an overall factor and so we construct a prime factorization of $ab$ that contains a $q$. But then the right hand side has only $p$ as prime factors. Since the two prime factorizations are the same, and factorizations are unique, this is impossible. Hence $a,b$ are formed entirely of $p$s in their prime decomposition.

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Prime Numbers

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