Closed form of sum of n^n series?

Question

Hi readers, may i know how this formula is derived? I thought of using https://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula which is Euler-Maclaurian to find the formula but its not.

Answer 1

How I think the summation was derived:$$\sum_{k=1}^n k^k= 1^1 + 2^2 + 3^3+....+n^n$$ $$=n^n\left(1+ \frac{(n-1)^{n-1}}{n^n}+\frac{(n-2)^{n-2}}{n^n}+\frac{(n-3)^{n-3}}{n^n}+...+\frac{1^{1}}{n^n}\right)$$ $$=n^n\left(1+ \frac{(n-1)^{n-1}}{n^{n-1}\cdot n}+\frac{(n-2)^{n-2}}{n^{n-2}\cdot n^2}+\frac{(n-3)^{n-3}}{n^{n-3}\cdot n^3}+...+\frac{1^{1}}{n^1\cdot n^{n-1}}\right)$$ $$=n^n\left(1+ \frac{1}{n}\left(\frac{n-1}{n}\right)^{n-1}+\frac{1}{n^2}\left(\frac{n-2}{n}\right)^{n-2}+\frac{1}{n^3}\left(\frac{n-3}{n}\right)^{n-3}+...+\frac{1}{n^{n-1}}\left(\frac{1}{n}\right)^{1}\right)$$