In Rudin's Principle of Mathematical Analysis (Indian Subcontinent Reprint of $3^{rd}$ Edition), while proving that $ (1+\frac{1}{n})^n = e $ (Theorem 3.31) Rudin states that:
\begin{eqnarray} \left(1+\frac{1}{n} \right)^n & = & 1+1+\frac{1}{2!}\left(1 - \frac{1}{n}\right) + \frac{1}{3!}\left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right) +\cdots+\frac 1 {n!}\left(1-\frac 1 n\right)\left(1-\frac 2 n\right)\cdots \left(1-\frac {n-1} n\right) \\ &=& \sum_{i=0}^{n}\frac{1}{i!}\left(\prod_{j=0}^{i-1}\left(1 - \frac{j}{n}\right)\right) \end{eqnarray}
Rudin writes that this is expression comes from using Binomial Theorem, but does not elaborate further. Applying the Binomial Theorem myself, I get \begin{eqnarray} \left(1+\frac{1}{n} \right)^n &=& 1 + {n \choose 1} \frac{1}{n} + {n \choose 2} { \left( \frac{1}{n} \right)}^2 + \cdots \\ &=& \sum_{i=0}^{n}{n \choose i} 1^{n-i} \left(\frac{1}{n}\right)^{i} \\ &=& \sum_{i=0}^{n}\frac{n!}{i!(n-i)!}\left(\frac{1}{n}\right)^i \end{eqnarray} What next? How to proceed further?