$$\eqalign{ & {a_n} = \frac{1}{n}\prod\limits_{k = 1}^n {{e^{\frac{1}{k}}}} \cr & \log {a_n} = - \log n + \sum\limits_{k = 1}^n {\frac{1}{k}} \cr & \log {a_n} = \sum\limits_{k = 1}^n {\frac{1}{k}} - \log n \cr & \mathop {\lim }\limits_{n \to \infty } \log {a_n} = \gamma \cr} $$
You can prove $0 < \gamma < 1$ since we can replace $\log n$ by $\log (n+1)$ and put
$$\eqalign{ & {b_n} = \sum\limits_{k = 1}^n {\frac{1}{k}} - \sum\limits_{k = 1}^n {\log \frac{{k + 1}}{k}} \cr & {b_n} = \sum\limits_{k = 1}^n {\left( {\frac{1}{k} - \log \frac{{k + 1}}{k}} \right)} \cr} $$
Since we know $$1 - \frac{1}{x} \leqslant \log x \leqslant x - 1$$
We have
$$\frac{1}{{k + 1}} \leq \log \left( {1 + \frac{1}{k}} \right) \leq \frac{1}{k}$$
We can prove both bounds with this. $$\eqalign{ & \frac{1}{k} - \frac{1}{{k + 1}} \geqslant \frac{1}{k} - \log \left( {\frac{{k + 1}}{k}} \right) \geqslant 0 \cr & \sum\limits_{k = 1}^n {\frac{1}{k} - \frac{1}{{k + 1}}} \geqslant \sum\limits_{k = 1}^n {\frac{1}{k} - \log \left( {\frac{{k + 1}}{k}} \right)} \geqslant 0 \cr & 1 \geqslant \sum\limits_{k = 1}^n {\frac{1}{k} - \log \left( {\frac{{k + 1}}{k}} \right)} \geqslant 0 \cr} $$
The first inequality also proves each term is positive or zero (though the last is discarded by a simple look at the image), i.e
$$\frac{1}{k} - \log \left( {\frac{{k + 1}}{k}} \right) \geqslant 0$$
