For example, for n = 2, the non-empty subsets are {1}, {2}, {1,2}, and the sum is \frac{1}{1} + \frac{1}{2} + \frac{1}{1 \cdot 2} = 2.

I think mathematical induction works. If the statement is true for $n=k$, let us consider $n=k+1$. [Moderator edit: The $k$ here is not the $k$ from the original problem, but a new variable. -- Darij] All non-empty subsets from $\{1,2,\dots,k\}$ are subsets from $\{1,2,\dots,k+1\}$, and the sum over those subsets is $k$. We need to prove that the sum over the subsets which contain $k+1$ is 1. For a set $S$ which contains $k+1$, $S\setminus\{k+1\}$ is a subset of $\{1,2,\dots,k\}$. Those kind of sets consist of (i)all non-empty subsets from $\{1,2,\dots,k\}$ and (ii)an empty set. Sum over part (i) is $k/(k+1)$, and part (ii) adds $1/(k+1)$ to the sum. thus the statement is true of $n=k+1$.

Another approach is to expand the product $\prod_{i = 1}^n (1 + \frac{1}{i})$.