Divisor Sum of Power of Prime

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

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

That is, let $$\sigma \left({n}\right)$$ be the sum of all positive divisors of $$n$$.

Then $$\sigma \left({n}\right) = \frac {p^{k+1} - 1} {p - 1}$$.

Proof
The divisors of $$n = p^k$$ are $$1, p, p^2, \ldots, p^{k-1}, p^k$$.

Hence from Sum of Geometric Progression $$\sigma \left({p^k}\right) = 1 + p + p^2 + \cdots + p^{k-1} + p^k = \frac {p^{k+1} - 1} {p - 1}$$.