Power of Prime is Deficient

Theorem
Let $n \in \Z_{>0}$ be a power of a prime number $p$:
 * $n = p^k$

for some $k \in \Z_{>0}$.

Then $n$ is deficient.

Proof
From Sigma Function of Power of Prime:
 * $\map \sigma n = \dfrac {p^{k + 1} - 1} {p - 1}$

Thus:

If $p = 2$ then $\dfrac {p^k} {p - 1} - p^k = 0$ and so:
 * $\dfrac {p^k} {p - 1} - p^k - \dfrac 1 {p - 1} = - \dfrac 1 {p - 1} < 0$

If $p > 2$ then $\dfrac {p^k} {p - 1} < p^k$ and so:
 * $\dfrac {p^k} {p - 1} - p^k - \dfrac 1 {p - 1} < 0$

Thus in all cases:
 * $\map A n < 0$

and so $n = p^k$ is deficient by definition.