How to show that $ \left(1+\frac{1}{n} \right)^n = \sum_{i=0}^{n}\frac{1}{i!}\left(\prod_{j=0}^{i-1}\left(1 - \frac{j}{n}\right)\right)$ [duplicate]

I think you can continue from there: $$\begin{align}{n \choose k}\frac 1{n^{n-k}}&=\frac{n(n-1)(n-2)\cdots (n-k+1)}{k!n^{n-k}}\\ &=\frac{1}{k!n^{n-k}}\prod_{r=0}^{k-1}(n-r)\\ &=\frac{1}{k!}\prod_{r=0}^{k-1}\frac{n-r}{n}\end{align}$$

and you can make the final step!. But for the indexes for the sum in binomial theorem you might write: $$(1+\frac 1n)^{n} = \sum_{k=0}^n{n \choose k}1^k(\frac 1n)^{n-k}=1 + \sum_{k=1}^{n}{n \choose k}\frac 1{n^{n-k}}$$ (because as you know ${n \choose 0}=1$ and what I wrote in the product above is not be true id $k=0$ because in this case we will have $\prod_{r=1}^{0-1}$ which does not have a sense)