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 $p^{n-1} \ldots$

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.
