Divisor Count Function of Power of Prime

Theorem
Let $n = p^k$ be the power of a prime number $p$.

Let $\map {\sigma_0} n$ be the divisor counting function of $n$.

That is, let $\map {\sigma_0} n$ be the number of positive divisors of $n$.

Then:
 * $\map {\sigma_0} n = k + 1$

Proof
From Divisors of Power of Prime, the divisors of $n = p^k$ are:
 * $1, p, p^2, \ldots, p^{k - 1}, p^k$

There are $k + 1$ of them.

Hence the result.