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? So I've known this for a while, and only finally thought to ask about it.. so, any prime number ($p$) to a power $n$ has the factors $\{p^{n-1},\ p^{n-2},\ ...\ p^1,\ p^0 = 1\}$
So, e.g., $5^4 = 625$, its factors are:
$$
\{625 = 5^4,\ 125 = 5^3,\ 25 = 5^2,\ 5 = 5^1,\ 1 = 5^0\}
$$
Now, my best guess is that it's related to its prime factorisation, $5*5*5*5$, but other than that, I have no idea.
So my question is, why does a prime number raised to a power $n$ have only the factors of $p^{n-1}$ and so on like above? No, I'm not talking about prime factorisation, I'm talking about normal factors (like 12's factors are 1, 2, 3, 4, 6, 12) and why $p^n$ doesn't have any other (normal) factors other than $p^{n-1} \ldots$
 A: 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.
A: 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.
A: The Fundamental theorem of arithmetic states that any $n \in \mathbb{Z}$ can be written in the form
$$n = \prod_{i = 0}^k p_i^{m_i}$$
for any primes $p_i$ and integers $k, m_i$, and each representation is unique for each $n$. What this basically means is that any $n$ can be split into a product of its primes.
If we have some $n = p^k$, then simply, its primary decomposition is just going to be $\underbrace{p \times p \times \cdots \times p}_k$.
A: You have to remember that factors of a number are just products of different combinations of its prime factors. For example, $36$ has prime factorization $2*2*3*3$. Now not only are 2 and 3 factors of 36, but also any product combination of them. So for example, $2*3=6$ is a factor, $2*3*3=18$ is a factor, and so on. So you know that a prime number has no factors other than 1 and itself, and a prime number to a power is just $p*p*p*...*p$. Let's use $5^4$ as an example. $5^4=5*5*5*5$. If we use our grouping technique of finding factors from above, the only combinations we can make from the factors are $5$, $5*5$, $5*5*5$ which correspond to $5^1$, $5^2$, and $5^3$ respectively.
A: A number $q > 1$ is prime by (a) definition if $q|ab$ implies $q|a$ or $q|b$. Thus if $q|p^n$, then $q|p$ or $q|p^{n-1}$. Now induct on $n$.
