=== Rough, Informal Proof ===
In brief: To maximize the product of a set of numbers, S, who's sum is an arbitrary integer > 1, n, two rules must be followed:
  1.  The set must contain some combination of 2s and 3s (but not necessarily both).
  2.  The number of 3s should be maximized while not violating 1.

For any n, n%3 (where % is the [http://en.wikipedia.org/wiki/Modulo_operation modulo operation]) will either be 0, 1 or 2. Case n%3:
  0: The maximum product of set S will be 3^(n/3) (set S is all threes)
  1: The maximum product of set S will be 3^((n-4)/3)*4 (set S has two twos and the rest are threes)
  2: The maximum product of set S will be 3^((n-2)/3)*2 (set S has one two and the rest are threes)

Proof:
The key to this problem is maximizing the function f(x) = x^(1/x).

Because this is '''not''' 137, a graphing calculator comes in very handy:<br>
http://www.cdf.toronto.edu/~c6woleve/x1overx.png
<br>
This means that the maximum must be somewhere between x=2 and x=4.  Solving for each gives:
  2^(1/2) = 1.4142
  3^(1/3) = 1.4423
  4^(1/4) = 1.4142
Because f(x) = x^(1/x) is maximized when x=3 (ok, well, the actual maximum is around 2.7, but we can only work with integers), we try to maximize the number of 3s in set S.  Because 2 is the second largest (keeping in mind that 4 = 2*2), after maximizing the number of 3s, the number of 2s is maximized.

Andrew and Tin (please forgive me if I've gotten your name wrong -_-') were in the group that figured this out. I just wrote up the "proof" (if it can really be called a proof).