Below is another approach:
The value of $1 invested for n years at interest rate i, compounded x times per year is (1+(i/x))nx.
In this case f(x) = the value of $1 invested at 100% interest compounded n times per year = (1+(1/x))x, where x approaches infinity.
The following table shows values of f(x) for various values of x:
x f(x)
----------- ----------
1 2.00000000
4 2.44140625
12 2.61303529
52 2.69259695
365 2.71456748
1,000 2.71692393
10,000 2.71814593
100,000 2.71826824
1,000,000 2.71828047
10,000,000 2.71828169
100,000,000 2.71828180
As x approaches infinity it can be seen that f(x) approaches e =~ 2.71828182846.
Michael Shackleford, A.S.A.