How to prove that $a_n =(1+\frac1n)^n$ for all $n\ge 1$ applies to $a_n=\sum_{k=0}^n \frac1{k!}(1-\frac1n)(1-\frac2n)\cdots(1-\frac{k-1}n)$?

Question

In an assignment for the course Real-Analysis I have to show that for all $n \ge 1$ applies: $$a_n = \displaystyle \sum_{k=0}^n \frac1{k!}\left(1-\frac1n\right)\left(1-\frac2n\right)\cdots\left(1-\frac{k-1}n\right)$$ where $a_n$ is given as: $$a_n =\left(1+\frac1n\right)^n$$

I have to use the binomial formula for this: $(x+y)^n=\displaystyle\sum_{k=0}^n \binom{n}{k} x^{k} y^{n-k}$

However, me and my fellow students have been trying to figure it out for some time now, but we can't...

Answer 1

$$LHS = \sum_{k=0}^n \frac{1}{k!} \prod_{j=0}^{k-1}\left(1-\frac{j}{n} \right)$$

$$= \sum_{k=0}^n \frac{1}{k!} \frac{1}{n^{k-1}} \prod_{j=0}^{k-1}\left(n- j \right)$$

$$= \sum_{k=0}^n \frac{1}{n^{k-1}} \left(\frac{1}{k!}\prod_{j=0}^{k-1}\left(n- j \right) \right)$$

Note that $$\frac{1}{k!}\prod_{j=0}^{k-1}\left(n- j \right) = \binom {n}{k}$$

Hence $$LHS = \sum_{k=0}^n \binom {n}{k} \frac{1}{n^{k-1}}$$

$$= \left(1+\frac{1}{n} \right)^n$$