Question

How can I verify if this is true for \(n\in\mathbb N\)?
\[\Large \dfrac1{n^n}=\sum_{i=0}^n\sum_{k=0}^i\dfrac{(-1)^kn^{k-i}}{k!(n-i)!}\]

Answer 1

It is true for \(n=1\) and \(n=2\).

Answer 2

This site is terrible \[\Large \dfrac1{n^n}=\sum_{i=0}^n\sum_{k=0}^i\dfrac{(-1)^kn^{k-i}}{k!(n-i)!}\]

Answer 3

yes @Kainui i completely agree :P

Answer 4

I stumped this while solving big kid club lol @Kainui

Answer 5

wait no lol i read ur question in some other way :P
sorry its not zeta

Answer 6

seems to be true
http://jsfiddle.net/ganeshie8/ykrpkweu/

Answer 7

is it enuff if i show it to be true for the integral analog of it