Sum of the series $\sum_{i=1}^n\ (n+1-i)^i \\ $ I am trying to find the sum of $\sum_{i=1}^n  (n+1-i)^i  $
The series also can be represented as $\sum_{i=1}^n i^{n+1-i} $
It seems like a very difficult nut to crack. Attempting this has taught me a lot about other series but I have made no real headway here. 
I considered manipulating
$S = n^1+(n-1)^2+...+1^n$
$S-(1+n)=(n-1)^2+...+2^{n-1}$
$n*S=n^2+n[(n-1)^2+...+2^{n-1}]+n$
$n*S=n^2+n[S-(1+n)]+n$
$n*S=n*S$
 A: Not an answer, but it would be too long for a comment
Basically your series, if we take the very first terms, is:
$$n + (n-1)^2 + (n-2)^3 + (n-3)^4 + \cdots $$
And we can write it also in this easier way:
$$\sum_{k = 0}^N (N-k)^{k+1} = \sum_{k = 0}^N (N-k)^k\cdot (N-k)$$
I checked with Mathematica and it doesn't spit out anything.. So I don't think a close form does exist.
Eventually we may split the two sum as
$$N\ \sum_{k = 0}^N (N-k)^{k} - \sum_{k = 0}^N k(N-k)^k$$
But no close forms do exist also for these ones.
You may try to evaluate some terms or to take some tries with respect to a chosen $N$. Probably a close form does not exist because it's a too irregular series.
However, the very fist computation for the series $S$ are (may be useful for you or someone else):
$$N = 2 ~~~ \to ~~~ S = 3$$
$$N = 3 ~~~ \to ~~~ S = 8$$
$$N = 4 ~~~ \to ~~~ S = 22$$
$$N = 5 ~~~ \to ~~~ S = 65$$
$$N = 6 ~~~ \to ~~~ S = 209$$
$$N = 7 ~~~ \to ~~~ S = 732$$
$$N = 8 ~~~ \to ~~~ S = 2780$$
$$N = 9 ~~~ \to ~~~ S = 11377$$
$$N = 10 ~~~ \to ~~~ S = 49863$$
Notice that the $N-th$ coefficient will always be zero, indeed it is
$$(N-N)^{N+1} = 0^{N+1}$$
Whilst the $(n-1)$-th coefficient will always be one:
$$(N - (N-1))^{N-1+1} = 1^N$$
And the $(N-2)$-th coefficient will be
$$(N - (N-2))^{N-2+1} = 2^{N-1}$$
And so on..
