I understand how to prove this problem, essentially the middle term $\sum_{j=0}^n \frac{1}{j!}$ is equal to the Euler's number, e, and the third term in this sequence is equal to 3. However, I am not entirely sure how I would show using just the binomial theorem that all of this is true.
$(1+\frac{1}{n})^n < \sum_{j=0}^n \frac{1}{j!} < 2 + \frac{1}{2} + \frac{1}{4} + ...+ \frac{1}{2^{n-1}}$
You would use binomial theorem in expressing your first term as a sum.
So you have $$(1+\frac{1}{n})^n = \sum_{i=0}^n{ \frac{1}{n^i}{ n \choose i }}$$ which you can use with the fact that $${ n \choose k } < \frac{n^k}{k!}$$, to get $$\sum_{i=0}^n{ \frac{1}{n^i}{ n \choose i }} < \sum_{i=0}^n{ \frac{1}{i!}}$$.