Sum with many troubles I am currently considering a sum $$\sum_{r=0}^{n}{\binom{n}{r} (-1)^{r} (1-\frac{r}{n})^{n}}$$ but have no thoughtful ideas how to start. 
Maybe it's worth noticing that $$\sum_{r=0}^{n}{\binom{n}{n-r}(-1)^{r}(1-\frac{r}{n})^{n}} = \sum_{r=0}^{n}{\binom{n}{r} (-1)^{r} (1-\frac{r}{n})^{n}}$$ and apply to the first one the convolution rule, but this doesn't seem productive enough.
Are there any hints that might help? Would be grateful to recieve any.
 A: We'll prov it combinatorically.
First note that $$\sum_{r=0}^{n} {n\choose r} (-1)^r (1-\frac rn)^n = \frac1{n^n} \sum_{r=0}^{n} {n\choose r} (-1)^r (n-r)^n$$
Now consider the rhs.
By Inclusion exclusion principle, we can see that the sum is exactly the number of words in length $n$ with the letters $\{1,...n\}$ such that each letter appears at least once:
if we consider $p_i = $the letter i doesnt appear, we have:
$W(p_{i_1},....,p_{i_r})= (n-r)^n$ (the letters $i_1,...,i_r$ doesnt appear, so we pick for each of the $n$ spots of the word a letter out of the $n-r$ letters remaining).
Then $$W(r) = {n \choose r}(n-r)^n$$
And finally we want $$E(0) = \sum_{r=0}^{n} {n\choose r} (-1)^r W(r)$$
Which combinatorical meaning is that each letter appears at least once. 
So the rhs sum is exactly the number of words of size $n$ that each letter appear exactly once, which means the permutations over $n$ different letters, which is $n!$
Therefore you get that the lhs is equal $$\frac{n!}{n^n}$$
A: The construction 
$$
\sum_{k=0}^n\binom{n}{k}(-1)^k f(x-kh)
$$
is the $n$-th iterate of the difference operator $(\Delta_h f)(x)=f(x)-f(x-h)$. Applied to a polynomial $f(x)=a_nx^n+...+a_1x+a_0$ of degree $n$ all terms vanish except the leading one which contributes its coefficient times $n!\,h^n$, $(Δ_h^{\,n}f)(x)=n!\,a_nh^n$.
In the question, $f(x)=x^n$, $h=1/n$ and the final evaluation is at $x=1$.
A: This  one can  also  be done  using  complex variables.

Suppose we are trying to evaluate
$$\sum_{r=0}^n {n\choose r} (-1)^r 
\left(1-\frac{r}{n}\right)^n.$$
Introduce the integral representation
$$\left(1-\frac{r}{n}\right)^n
= \frac{n!}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1}{z^{n+1}} 
\exp\left(\left(1-\frac{r}{n}\right)z\right) \; dz.$$
This gives for the sum
$$\frac{n!}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1}{z^{n+1}} 
\sum_{r=0}^n {n\choose r} (-1)^r
\exp\left(\left(1-\frac{r}{n}\right)z\right) \; dz
\\ = \frac{n!}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1}{z^{n+1}} \exp(z)
\sum_{r=0}^n {n\choose r} (-1)^r
\exp\left(-\frac{r}{n}z\right)\; dz
\\ = \frac{n!}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1}{z^{n+1}} \exp(z)
\left(1-\exp(-z/n)\right)^n \; dz.$$
Therefore the sum is
$$n! [z^n] \exp(z) \left(1-\exp(-z/n)\right)^n.$$
Now $1-\exp(-z/n)$ starts at $z/n$ so the only contribution is
$1/n^n$ for a final answer of
$$\frac{n!}{n^n}.$$
