Is $\sum_{k=1}^{n} k^{k-1} (n-k)^{2n-k} \binom{n}{k} \sim\frac{n^{2n}}{2\pi} $? How can you compute the asymptotics of
$$T=\sum_{k=1}^{n} k^{k-1} (n-k)^{2n-k} \binom{n}{k}\;?$$
This is related to Asymptotics of sum of binomials .
I attempted to simply use Stirling's approximation. This gives you
$$T \approx U =\frac{n^{n+1/2}}{\sqrt{2\pi}}\sum_{k=1}^n \frac{(n-k)^{n-1/2}}{k^{3/2}}$$
This approximation $U$ seems to be some constant times $n^{2n}$ asymptotically and also for $T$ itself.  
By experimentation it looks like 
$$T \sim \frac{n^{2n}}{2\pi}.$$
Is this correct?
Update. No it isn't.
 A: The summand is decreasing with $k$:
$$
  \sum_{k=1}^n k^{k-1} (n-k)^{2n-k} \binom{n}{k} = n^{2n} \sum_{k=1}^n \left(\frac{k}{n}\right)^{k-1} \left(1-\frac{k}{n}\right)^{2n-k} \frac{1}{n} \binom{n}{k}
$$
We can now use the fact that 
$$
   \lim_{n \to \infty} \left(\frac{k}{n}\right)^{k-1} \left(1-\frac{k}{n}\right)^{2n-k} \frac{1}{n} \binom{n}{k} = \frac{k^{k-1}}{k!} \left( \frac{1}{\mathrm{e}^2} \right)^k
$$
Thus
$$
 \sum_{k=1}^n k^{k-1} (n-k)^{2n-k} \binom{n}{k}  \rightarrow_{n \to \infty} n^{2n} \sum_{k=1}^\infty \frac{k^{k-1}}{k!} \left( \frac{1}{\mathrm{e}^2} \right)^k = n^{2n} \left( -W\left(-\frac{1}{\mathrm{e}^2}\right)\right) = n^{2n} \exp\left( W\left(-\mathrm{e}^{-2}\right) - 2 \right)
$$
where $W(z)$ denotes the Lambert $W$-function. Note that 
$$
   \left( -W\left(-\frac{1}{\mathrm{e}^2}\right)\right) \approx 0.158594 < 0.159155 \approx \frac{1}{2\pi}
$$
Numerically it seems that estimate is from above:

A: Since the terms in the sum are decaying very fast, we can restrict the sum to small values of $k$ and use an asymptotic estimate that has a small error for large $n$ and small $k$. In more detail, if
$$a_k = \binom n k k^{k-1} (n-k)^{2n-k},$$
then
$$\frac {a_{k+1}} {a_k} =
\left( 1 + \frac 1 k \right)^{k-1}
 \left( 1 - \frac 1 {n-k} \right)^{2n-2k}
 \left( 1 - \frac 1 {n-k} \right)^{k-1} < \frac 1 e, \\
\sum_{k = \ln^2 n + 1}^n a_k < a_1 \sum_{k = \ln^2 n}^{n - 1} e^{-k} =
O \left( n^{2n} e^{-\ln^2 n} \right),$$
thus the tail will be negligible compared to all the terms in the expansion $n^{2n} \left( c_0 + \frac {c_1} n + \dots \right)$. On the other hand, for large $n$,
$$a_k = \frac {k^{k-1} e^{-2 k}} {k!} n^{2 n}
 \left( 1 - \frac {k^2 - k} {2 n} +
  O \left(\frac {k^4} {n^2} \right) \right), \\
\sum_{k=1}^{\ln^2 n} a_k =
n^{2 n} \left(
 \sum_{k=1}^{\ln^2 n} \frac {k^{k-1} e^{-2 k}} {k!}
   \left( 1 - \frac {k^2 - k} {2 n} \right) +
 O \left(\frac {\ln^{10} n} {n^2} \right) \right),$$
once we prove that the $O$ term is uniform in $k$.
Using the same reasoning, we can now extend the summation range  on the right-hand side to infinity, as the contribution from the tail is $O(e^{-\ln^2 n})$. This yields the Lambert W-function. The second term contains the factors $k^p$, which correspond to applying the operators $\left( z \frac d {dz} \right)^p$ to the W-function. After simplifications,
$$\sum_{k=1}^n \binom n k k^{k - 1} (n - k)^{2 n - k} \sim
n^{2 n} \left(
 \alpha - \frac {\alpha^2 (2 - \alpha)} {2(1 - \alpha)^3 n} + \dots \right), \\
\alpha = -W \!\left(-e^{-2} \right).$$
