<s>[INST] Sum of powers-of-integers bound 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: $1^4 + 2^4 + 3^4 = 1+16+81 = 98 < 4^4 = 256.$
 [/INST] For all $0 \leqslant r \leqslant n-1$, we have
$$
\frac{(r+1)^n}{r^n} = \left( \frac{r+1}{r} \right)^n =  \left( 1 + \frac{1}{r} \right)^n \geqslant 1 + \frac{n}{r} \geqslant 2,
$$
so $r^n \leqslant \frac{1}{2} (r+1)^n$. Now by a simple induction on $k$, we obtain
$$
(n-k)^n \leqslant \frac{n^n}{2^k}.
$$
Summing over $k = 1, 2, \ldots, n-1$, we get
$$
\sum_{k=1}^{n-1} (n-k)^n \leqslant n^n \sum_{k=1}^{n-1} 2^{-k} < n^n \sum_{k=1}^{\infty} 2^{-k} = n^n,
$$
which is what we wanted to show. 

This approach can squeeze out a little more. In the very first step, assuming $\varepsilon > 0$ and $n$ sufficiently large, we have
$$
\frac{(r+1)^n}{r^n} =  \left( 1 + \frac{1}{r} \right)^n \geqslant \left( 1 + \frac{1}{n} \right)^n  \geqslant \mathrm e - \varepsilon.
$$
Plugging in this refined bound in the above proof, we get the upper bound
$$
\sum_{k=1}^{n-1} k^n \lt n^n \cdot \sum_{k=1}^{\infty} (\mathrm{e} - \varepsilon)^{-k} = \frac{n^n}{\mathrm e - \varepsilon - 1}.
$$
This gives
$$
\limsup_n  \ n^{-n} \cdot  \sum_{k=1}^{n-1} k^n \leqslant \frac{1}{\mathrm e  - 1},
$$
that matches (one half of) the limit obtained by Mike Spivey. 
</s>