sequences and series - Compute $lim_{n to +infty} n^{-frac12 left(1+frac{1}{n}right)} left(1^1 cdot 2^2 cdot 3^3 cdots n^n right)^{frac{1}{n^2}}$

How to compute
$$\displaystyle \lim_{n \to +\infty} n^{-\dfrac12 \left(1+\dfrac{1}{n}\right)} \left(1^1\cdot 2^2 \cdot 3^3 \cdots n^n \right)^{\dfrac{1}{n^2}}$$
I'm interested in more ways of computing limit for this expression

My proof:

Let $u_n$be that sequence we've:

$$\begin{eqnarray*} \ln u_n &=& -\frac{n+1}{2n}\ln n + \frac{1}{n^2}\sum_{k=1}^n k\ln k\\ &=& -\frac{n+1}{2n}\ln n + \frac{1}{n^2}\sum_{k=1}^n k\ln \frac{k}{n}+\frac{1}{n^2}\sum_{k=1}^n k\ln n\\ &=& \frac{1}{n^2}\sum_{k=1}^n k\ln \frac{k}{n}\\ &=& \frac{1}{n}\sum_{k=1}^n \frac{k}{n}\ln \frac{k}{n}\\ &\to&\int_0^1 x\ln x\,dx = -1/4 \end{eqnarray*}$$

Therefore the limit is $e^{-\frac{1}{4}}$