$\underset{n\to \infty }{\lim }{n}^{-\frac{1}{2}(1+\frac{1}{n})}{(1^1\cdot 2^2\cdot \cdot \cdot \cdot n^n)}^{\frac{1}{n^2}}=e^{-1/k},k\in N,$ then the number of proper divisor(s) of ‘k’ will be____

$$\begin{gathered} \underset{n\to \infty }{\lim }{n}^{-\frac{1}{2}(1+\frac{1}{n})}{(1^1\cdot 2^2\cdot \cdot \cdot \cdot n^n)}^{\frac{1}{n^2}} \\ =\underset{n\to \infty }{\lim }\frac{{(1^1\cdot 2^2\cdot \cdot \cdot \cdot n^n)}^{\frac{1}{n^2}}}{n^{\frac{1}{2n}(n+1)}}=\underset{n\to \infty }{\lim }{\left(\frac{1^1\cdot 2^2\cdot \cdot \cdot \cdot n^n}{n^{{\displaystyle \frac{n\left(n+1\right)}{2}}}}\right)}^{\frac{1}{n^2}} \\ L=\underset{n\to \infty }{\lim }{\left(\frac{1}{n}{\left(\frac{2}{n}\right)}^2.{\left(\frac{3}{n}\right)}^3....{\left(\frac{n}{n}\right)}^n\right)}^{\frac{1}{n^2}} \\ L=\underset{n\to \infty }{\lim }{\left({\left(\frac{1}{n}\right)}^{\frac{1}{n}}{\left(\frac{2}{n}\right)}^{\frac{2}{n}}.{\left(\frac{3}{n}\right)}^{\frac{3}{n}}....{\left(\frac{n}{n}\right)}^{\frac{n}{n}}\right)}^{\frac{1}{n}} \end{gathered}$$

$\log L=\underset{n\to \infty }{\lim }\frac{1}{n}\sum _{r=1}^n\log {\left(\frac{r}{n}\right)}^{\frac{r}{n}}$

$$\begin{gathered} \log L={\int }_0^1\ln x^{x}dx \\ \log L={\int }_0^{1}x\left(\ln x\right)dx \\ \log L={\left(\ln x\left(\frac{x^2}{2}\right)\right)}_0^1-{\int }_0^1\frac{1}{x}\left(\frac{x^2}{2}\right)dx \\ \log L=0-\frac{1}{2}{\int }_0^{1}xdx \\ \log L=\frac{-1}{4} \\ \Rightarrow L=e^{-1/4} \end{gathered}$$

divisors of k=4 are 1,2,4

proper divisor of k is 1,2

number of proper divisors=2