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:
 * $\sigma \left({n}\right) = \dfrac {p^{k+1} - 1} {p - 1}$

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

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