For each $n>0$, how do we prove that $$\Gamma'(n+1)> \log{n} \cdot \Gamma(n+1)$$
I had spent about half an hour on this question, but just could find any way of proceeding for the solution.
Wikipedia page gave me an interesting identity $$\Gamma'(n+1)= n! \cdot \Biggl( - \gamma + \sum\limits_{k=1}^{n} \frac{1}{k}\Biggr)$$ But i don't know how it can be applied here.
Well, if we rearrange the inequality to
$$\frac{\Gamma^{\prime}(n+1)}{\Gamma(n+1)} > \log n$$
and then note that
$$\frac{\Gamma^{\prime}(n+1)}{\Gamma(n+1)}=-\gamma+\sum_{j=1}^n \frac1{j}$$
you should be able to easily deduce the inequality.