Divisor Count Function of Power of Prime

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Theorem

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

Let $\map {\sigma_0} n$ be the divisor count 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.

$\blacksquare$


Sources