TITLE: Sum of powers-of-integers bound
QUESTION [7 upvotes]: I am looking for a nice proof of this inequality:
\begin{equation}
    \sum_{k=1}^{n-1} k^n < n^n, \quad n > 0.
\end{equation}
Example: $1^4 + 2^4 + 3^4 = 1+16+81 = 98 < 4^4 = 256.$

REPLY [4 votes]: The Mean Value Theorem says that for some $\xi$ between $k$ and $k+1$,
$$
(k+1)^{n+1}-k^{n+1}=(n+1)\xi^n>(n+1)k^n
$$
Therefore, by telescoping series,
$$
\sum_{k=0}^{n-1}(n+1)k^n<n^{n+1}
$$
Dividing by $n+1$, we get
$$
\begin{align}
\sum_{k=0}^{n-1}k^n
&<\frac{1}{n+1}n^{n+1}\\
&<n^n
\end{align}
$$