Other numbers with property $1 + 2 + \cdots + 2^{n-1} = 2^n - 1 $ It is well known that $$1 + 2 + \cdots + 2^{n-1} = 2^n - 1 $$
This means that $2^n$ is equal to one more than the sum of all its proper divisors.
It turns out the only integers with this property are of the form $2^n$ ($n \ge 0$), but I have no idea how to prove this. Any hints?
 A: Okay.  for $p$ prime.  $p \ne 2$, $1 + p + ... + p^{n} = \frac {p^{n+1} - 1}{p - 1} > p^n$.
So if $p^n$ is a factor the sum of factors will include summands $1, p, p^2.... p^n$ which when multiplied and distributed by other prime power factors will will be greater than $p^{n+1}*\text {sum of factors of m divided by p to the n}$ 
So if $m = \prod p_i^{k_i}$, with distribution and induction, sum of all factors > $\prod p_i^{k_i + 1}$ so, 1 more than sum of proper fractions $> \prod p_i^{k_i + 1} - m + 1 = \prod p_i^{k_i + 1} - \prod p_i^{k_i} + 1> \prod p_i^{k_i} = m$.
Okay, it needs to be polished and cases of m = p or multiple of single powers of primes need to be ironed out.  But in general that's the gyst.
======
Okay, more formally.
Let f(m) = sum of all factors (including m) of integer m.
Our property m = 1 + sum of proper factors $\iff$ f(m) = 2m - 1.
If $m = \prod p_i^{k_i}$ ($p_i$ prime). $f(p_i^{k_i}) = 1 + p_i + ... + p_i^{k_i} = \frac{p_i^{k_i + 1} - 1}{p_i - 1}$.  So $f(m) = \frac{p_i^{k_i + 1} - 1}{p_i - 1}*f(m/p_i^{k_i}) = \prod \frac{p_i^{k_i + 1} - 1}{p_i - 1} $
For all $p_i > 2$, $\frac{p_i^{k_i + 1} - 1}{p_i - 1} > p^{k_i}+1 $. 
So if $f(m = 2^n\prod p_i^{k_i}; p_i \text { odd}) = (2^{n+1} - 1)\prod \frac{p_i^{k_i + 1} - 1}{p_i - 1} > (2^{n+1} - 1)\prod (p_i^{k_i}+1)> 2m -1$ If any $p_i \ne 1$ and $n > 1$ or $n = 0$ but there are at least two $p_i$.
If, on the other hand, $m = p;p > 3$ then $ f(m) = 1 + p \ne 2p - 1$
So $m = 2^n$ are the only integers with this property.
