Tau of Power of Prime

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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.

$\blacksquare$