Let $(X_{i})_{i}$ be IID standard normal random variables. Define $M_{n}:=\max{(X_{1},...,X_{n})}$ and $a_{n}=\sqrt{2\log(n)-\log(\log{(n)})-\log{(4\pi)}}$
Show that $a_{n}M_{n} - a_{n}^{2}\xrightarrow{d} Q$ where $Q$ is a probability measure on $\mathbb R$ with the distribution function $F_{Q}(c)=\exp{(-e^{-c}})$ for $c \in \mathbb R$.
My ideas:
Let $c$ be an arbitrary continuous point for $F_{a_{n}M_{n} - a_{n}^{2}}$ and $F_{Q}$ respectively.
It follows that $F_{a_{n}M_{n} - a_{n}^{2}}(c)=P(a_{n}M_{n} - a_{n}^{2}\leq c)=P(M_{n}\leq\frac{c+a_{n}^{2}}{a_{n}})=P(X_{1}\leq \frac{c+a_{n}^{2}}{a_{n}},...,X_{n}\leq \frac{c+a_{n}^{2}}{a_{n}})=(P(X_{1}\leq \frac{c+a_{n}^{2}}{a_{n}}))^{n}=\phi(\frac{c+a_{n}^{2}}{a_{n}})^{n}=\phi(\frac{c+2\log(n)-\log(\log{(n)})-\log{(4\pi)}}{\sqrt{2\log(n)-\log(\log{(n)})-\log{(4\pi)}}})^{n}=\frac{1}{\sqrt{2\pi}}(\int_{-\infty}^{\frac{c+2\log(n)-\log(\log{(n)})-\log{(4\pi)}}{\sqrt{2\log(n)-\log(\log{(n)})-\log{(4\pi)}}}}e^{-\frac{x^2}{2}}dx)^{n}$
And then I get stuck.
See e.g. the lecture notes of Anton Boviar on “Extreme values of random processes” on pp. 5-7.