Prove the following limit about $e$ I need to show that: $$\lim_{n \to +\infty}\sum_{k=0}^{k=n-1}\left(\frac{n-k}{n}\right)^n = \frac{e}{e-1}$$I observed that taking the limit term by term gives the result, but of course this is not justified. This exercise assume no prior knowledge on uniform convergence of series. (I am new to series but I remember from calculus a result on limits that is justified in this case.) So I should be able to prove it following another path. 
It also appears that a conversion to a definite integral is unfeasible. What do you suggest? 
 A: COMMENT.- Except for some trick of which I do not believe, we must first find the form of $$S_n=\sum_{k=0}^{k=n-1}\left(\frac{n-k}{n}\right)^n$$ which is equal to $$S_n=1+\frac{1}{n^n}\left(1+2^n+3^n+\cdots+(n-1)^n\right)$$
So you have a problem that involves the sum of the $n-1$ first $n-\text {powers}$. Faulhaber's Formula gives
$$S_n=1+\frac{1}{n^n}\left(\frac{(n-1)^{n+1}}{n+1}+\frac12((n-1)^n+\sum_{k=2}^n\frac{B_k}{k!}\frac{n!}{(n-k+1)!}n^{n-k+1}\right)$$ where $B_k$ is the Bernoulli number of index $k$ so you can simplify something the expresion because $B_{2k+1}=0$ and you have to consider only the $B_{2k}$. You can still move forward by examining theory about Bernoulli numbers.In particular the Bernoulli polynomials are defined by
$$B_k(x)=\sum\binom nkB_{n-k}x^k$$ and the maybe important here is the generating function of these polynomials which is $$\frac{te^{tx}}{e^t-1}=\sum_{k=0}^{\infty}B_k(x)t^k$$
A: To summarize my comments, we can define for each positive integer $n$: 
$$ f_n(k) = (1-k/n)^n \quad \forall k \in \{0, 1, ..., n-1\}$$
and we observe that for any positive integer $m$: 
$$ \sum_{k=0}^{m-1} f_n(k) \leq \sum_{k=0}^{n-1} f_n(k) \leq \sum_{k=0}^{n-1} e^{-k} \quad \forall n>m$$
So we can take limits as $n\rightarrow\infty$.
