Limit of equation as $m \rightarrow \infty$

Question

I was taking a look at this question: Convergence in distribution to a limit

And I’m struggling with working out the limit given at the end of the solution.

Overall I can get half of the way there, but the bit I’m missing is how as $m \rightarrow \infty$ then $$e^{-mw} \left( 1 - \frac{w}{m} \right)^{-m^{2}} \rightarrow e^{w^{2}/2}$$

Any explanation of calculating the limit would be appreciated!

Answer 1

$$-mw-m^{2} \ln (1-\frac w m)= -mw -m^{2}(-\frac w m +\frac {w^{2}} {2m^{2}}+o(m^{-2}))$$ $$=-\frac {w^{2}} 2 +o(1) \to -\frac {w^{2}} 2$$ so the limit is $e^{-w^{2}/2}$