<s>[INST] Asymptotics of $1^n + 2^{n-1} + 3^{n-2} +\cdots + (n-1)^2 + n^1$ Suppose $n\in\mathbb{Z}$ and $n > 0$.
Let $$H_n = 1^n + 2^{n-1} + 3^{n-2} +\cdots + (n-1)^2 + n^1.$$
I would like to find a Big O bound for $H_n$.  A Big $\Theta$ result would be even better.
 [/INST] My approach is very much along the same line as that of leonbloy.
Let $f(k) = k^{n+1-k}$. Use Euler-Maclaurin formula:
$$
 \begin{eqnarray}
   \sum_{k=1}^n f(k) &=& \int_1^n f(x) \mathrm{d} x + \frac{1}{2} \left( f(1) + f(n) \right) + \sum_{m=1}^q \frac{B_{2m}}{(2m)!} \left( f^{(2m-1)}(n) - f^{(2m-1)}(1) \right) \\ 
   &&+  \frac{1}{(2q)!} \int_1^n B_{2q}(\{t\}) f^{(2q)}(t) \mathrm{d} t
  \end{eqnarray}
$$
Since $f(1) = 1$ and $f(n)=n$. Asymptotically,  $f^{(2m-1)}(1) \sim -n \cdot \log^{2m-1}(n)$ and $f^{(2m-1)}(n) \sim n^{2m-1}$.
Thus the main term comes from the integral. Let $x=1+(n-1)t$.
$$
   \int_1^n x^{n+1-x} \mathrm{d} x = (n-1) \int_0^1 \left( 1 + (n-1) t \right)^{t+n(1-t)} \,\, \mathrm{d} t
$$
The integrand is sharp-peaked with peak location at 
$$
    t_0 = \frac{1}{n-1} \left( \frac{n+1}{W((n+1) \, \mathrm{e})} -1 \right)
$$
Logarithm of the integrand, expanded in the vicinity of $t_0$:
$$
  \left( n + t - n t \right) \log\left(1 + (n-1) t \right) = (n+1) \left( w_n + \frac{1}{w_n} - 2 \right) - \frac{\sigma_n}{2} ( t-t_0)^2 + o((t-t_0)^2)
$$
where $w_n = W((n+1)\mathrm{e})$ and $\sigma_n = \frac{(n-1)^2 w_n ( w_n + 1)}{n+1}$.
Thus 
$$
    \sum_{k=1}^n k^{n+1-k} \sim  \exp\left( (n+1) \left( w_n + \frac{1}{w_n} - 2 \right) \right) \sqrt{\frac{2\pi (n+1)}{w_n (1+w_n)}}
$$
Here is the numerical simulation, showing the agreement on logarithmic scale:

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