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 $p^{n-1} \ldots$

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 $\underbrace{p \times p \times \cdots \times p}_k$.
