$\lim_{n\rightarrow \infty}n^{-\left(1+1/n\right)/2}\times \left(1^1\times 2^2\times 3^3\times\cdots\times n^n\right)^{1/{n^2}}$ 
Evaluate the limit
  $$
y=\lim_{n\rightarrow \infty}n^{-\left(1+1/n\right)/2}\times \left(1^1\times 2^2\times 3^3\times\cdots\times n^n\right)^{1/{n^2}}
$$

My Attempt:
When $n\rightarrow \infty$, then $n^{-\left(1+1/n\right)/2}\rightarrow n^{-1/2}$. So the limit becomes
$$
y=\lim_{n\rightarrow \infty}n^{-1/2}\times \left(1^1\times2^2\times3^3\times\cdots\times n^n\right)^{1/{n^2}}
$$
Taking $\ln()$ on both side, we get
$$
\begin{align}
y&=\lim_{n\rightarrow \infty}\ln\left\{n^{-1/2}\times\left(1^1\times2^2\times3^3\times\cdots\times n^n\right)^{1/{n^2}}\right\}\\
&=\lim_{n\rightarrow \infty}\left\{-\frac{1}{2}\ln(n)+\frac{1}{n^2}\ln\left(1^1\times2^2\times3^3\times\cdots\times n^n\right)\right\}
\end{align}
$$
How can I complete the solution from this point?
 A: As already commented by Lucian, $$\prod_{k=0}^n k^k=H(n)$$ which is the hyperfactorial function which also has a Stirling-like series $$H(n)=A e^{-\frac{n^2}{4}} n^{\frac{n^2}{2}+\frac{n}{2}+\frac{1}{12}} \left(1+\frac{1}{720
   n^2}-\frac{1433 }{7257600 n^4}+O\left(\left(\frac{1}{n}\right)^6\right)\right)$$ where $A$  is the Glaisher-Kinkelin constant. 
Using it, the expression you are dealing with has then an asymptotic expansion $$\frac{1}{\sqrt[4]{e}}+\frac{12 \log (A)+\log (n)}{12
   \sqrt[4]{e} n^2}+O\left(\left(\frac{1}{n}\right)^{7/2}\right)$$
A: In your last expression, the second term equals
$$
\frac{1}{n^2} \sum_{k=1}^n k\ln k
$$
(the term you dropped in your line "my try" does go to one, so this is correct). By using integrals to estimate the sum, we see that this equals $(1/2)\ln n -1/4 + o(1)$, so the original expression converges to $e^{-1/4}$.
To justify the evaluation of $\sum k\ln k$ as $\int_1^n x\ln x\, dx$, notice that the integrand has derivative $\lesssim \ln x$, so the mistake I'm making when I replace $k\ln k$ by the integral over $[k-1,k]$ is $\lesssim \ln k$ and thus the total error is $O(n\ln n)$.
