Here $m=\frac{1}{2}n(\log n + w(n))$. The full claim is that
$$\left(1-o(1)\right) n \exp \bigg(-\frac{2m}{n-2}\bigg) \ge (1-o(1)) e^{-w}$$
but am I'm having trouble seeing why.
Edit: \begin{align*} \left(1-o(1)\right) n \exp \bigg(-\frac{2m}{n-2}\bigg) &= \left(1-o(1)\right) n \exp \bigg(-\frac{n(\log n + w(n))}{n-2}\bigg) \\ &= \left(1-o(1)\right) n \exp \bigg(-\left(1+\frac{2}{n-2}\right)\left(\log n + w(n)\right)\bigg) \\ &= \left(1-o(1)\right) n \exp \bigg(-\log n\bigg)\exp \bigg(\frac{-2 \log n}{n-2}\bigg) \exp \bigg(-w(n)\bigg) \exp \bigg(-\frac{2 w(n)}{n-2}\bigg)\\ &= \left(1-o(1)\right) n n^{-1}n^{\frac{-2}{n-2}} \exp \bigg(-w(n)\bigg) \exp \bigg(-\frac{2 w(n)}{n-2}\bigg)\\ &= \left(1-o(1)\right) n^{\frac{-2}{n-2}} \exp \bigg(-w(n)\bigg) \exp \bigg(-\frac{2 w(n)}{n-2}\bigg)\\ \end{align*}
and this gives the desired result because $n^{\frac{-2}{n-2}} \ge 1$ and $\exp \bigg(-\frac{2 w(n)}{n-2}\bigg) \ge 1$ if $w =o(n)$?