It wouldn't be a very cool proof, but you can let $\displaystyle u_n=\log(1^12^2\cdots n^n)-\frac 12 n^2\log n-\frac 12 n\log n+\frac 14 n^2$ and check that $\displaystyle u_n-u_{n-1}=\frac{1}{12n}+O\left(\frac{1}{n^2}\right)$ (very simple but lengthy, or Maple can do it for you). Since $\displaystyle \sum_{k=2}^n\frac{1}{k} = \log n -\gamma+o(1)$ and $\displaystyle \sum_k\frac{1}{k^2}$ converges, we deduce $\displaystyle u_n=\sum_{k=2}^n (u_k-u_{k-1})=\frac{1}{12}\log n-C+o(1)$. Taking exponentials gives the result...