Trying to find a closed form for $\sum_{k=1}^nk^n$ Edited:
Cosider the following summation $$A=\sum_{k=1}^nk^n.$$
I try to find an upper bound for $A$ as follow:
I try to prove that $$\sum_{i=1}^ni^n\leq 2n^n$$ as follow, first I checked base cases that holds. then I suppose $$\sum_{i=1}^ki^k\leq 2k^k$$ now to prove for $n=k+1$, $$\sum_{i=1}^{k+1}i^{k+1}=\sum_{i=1}^{k}i^{k+1}\;\;+ (k+1)^{k+1}$$. At this step I get stuck, How I can relate $\sum_{i=1}^{k}i^{k+1}$ to our induction assumption?
 A: From a formal point of view, there is one since it is the definition of a particular generalized harmonic number
$$\sum_{i=1}^n i^n=H_n^{(-n)}$$
A: No, there is not — but good asymptotic approximations are known. This is OEIS A031971.
A: An asymptotic approximation may be obtained as follows:
\begin{align*}
\sum\limits_{k = 1}^n {k^n } & = \sum\limits_{k = 0}^{n - 1} {(n - k)^n }  = n^n \sum\limits_{k = 0}^{n - 1} {\left( {1 - \frac{k}{n}} \right)^n } \\ & = n^n \sum\limits_{k = 0}^{n - 1} {\exp \left( {n\log \left( {1 - \frac{k}{n}} \right)} \right)} \\ & = n^n \sum\limits_{k = 0}^{n - 1} {\exp \left( { - k - \frac{{k^2 }}{{2n}} + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right)} \right)} 
\\ &
 = n^n \left( {\sum\limits_{k = 0}^{n - 1} {e^{ - k} }  - \frac{1}{{2n}}\sum\limits_{k = 0}^{n - 1} {k^2 e^{ - k} }  + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right)} \right) \\ & = n^n \left( {\frac{e}{{e - 1}} - \frac{{e(e + 1)}}{{2(e - 1)^3 }}\frac{1}{n} + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right)} \right)\\ & = \frac{e}{{e - 1}} n^n \left( {1 - \frac{{e + 1}}{{2(e - 1)^2 }}\frac{1}{n} + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right)} \right).
\end{align*}
