An indefinite integral $\displaystyle\int_{0}^{1}{\frac{te^t}{(e^t-1)^{n}}\mathrm{d}t}$ was given. The problem asks to show that it converges. The solution states the following:
$\frac{te^t}{(e^t-1)^{n}}\to\frac{1}{t^{n-1}}$ for $t\to0^+$. Since $\displaystyle\int_{0}^{1}{\frac{1}{t^{n-1}}\mathrm{d}t}$ converges for $n-1<1$ integral $\displaystyle\int_{0}^{1}{\frac{te^t}{(e^t-1)^{n}}\mathrm{d}t}$ also converges for $n-1<1$ or $n<2$.
We can easily show, using L'Hopital, that $\displaystyle\lim_{t\to0^+}{\frac{\frac{te^t}{(e^t-1)^n}}{\frac{1}{t^{n-1}}}}=1$, so this solution is correct. But how to derive the fact that the function $\frac{te^t}{(e^t-1)}$ approaches $\frac{1}{t^{n-1}}$ from the left side at $t=0$?
The problem in the denominator is when $t\to 0^+$. Recall that $e^t\sim1+t$: $$\frac{te^t}{(e^t-1)^n}\sim \frac{te^t}{t^n}=\frac{e^t}{t^{n-1}} \quad \text{as} \quad t\to0 $$ Since, $e^t \to 1$ as $t$ approaches $0$, you conclude that: $$\frac{te^t}{(e^t-1)^n}\sim \frac{1}{t^{n-1}} $$