When you compound 100% annual interest over n periods of 1/n of a year, your money is multiplied by a\_n = (1 + 1/n)\^n. The number e, according to one definition, is the limit approached by this expression as n tends to infinity. You would like to know why this number e is also expressible as 1 + 1 + 1/2! + 1/3! + 1/4! + ...
Often the answer to this question is given via calculus and Taylor series, but I'm going to give a more elementary answer. The key is to rewrite a\_n by using the binomial formula. I'll write C(n,k) for the binomial coefficient n!/k!(n-k)!.
a\_n = C(n,0) + C(n,1)/n + C(n,2)/n\^2 + ... + C(n,k)/n\^k + ... + C(n,n)/n\^n.
After some simplification and rewriting, this is
a\_n = 1 + 1 + (1/2!)(1 - 1/n) + (1/3!)(1 - 1/n)(1 - 2/n) + ... + (1/k!)(1 - 1/n)(1 - 2/n)...(1 - (k-1)/n) + ... + (1/n!)(1 - 1/n)(1 - 2/n)...(1 - (n-1)/n)).
As n increases, two things happen to this expression. First, the number of terms increases indefinitely. And second, if you examine just the (k+1)st term for a fixed value of k, the factors 1 - 1/n, 1 - 2/n, ..., 1 - (k-1)/n, all tend to 1, so the term itself approaches 1/k!.
It is plausible therefore to assert that a\_n approaches the sum 1 + 1 + 1/2! + 1/3! + 1/4! + ... as n tends to infinity.
To turn this argument into a correct proof would take some additional work. We would need to show why the fact that there are an ever-increasing number of terms in a\_n doesn't change our conclusion. But in fact, it is possible to do this by showing that these terms are "small" in a precise sense I won't go into.
Edit. For those who do want a full proof, note first that in fact a\_n is bounded above by 1 + 1 + 1/2! + 1/3! + ... Moreover, as all the terms in a\_n are positive, the argument we've just given shows that lim inf a\_n >= 1 + 1 + 1/2! + ... + 1/k! for any fixed k we choose. That is enough to conclude that lim a\_n = 1 + 1 + 1/2! + 1/3! + ...