Divisor Count Function of Power of Prime

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

Let $\tau \left({n}\right)$ be the tau function of $n$.

That is, let $\tau \left({n}\right)$ be the number of positive divisors of $n$.

Then $\tau \left({n}\right) = k + 1$.

Proof
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.