I have to solve the following question:
For $Z_1,\ldots,Z_p$ i.i.d. with $N(0,1)$ standard Gaussian distribution and $\alpha >0$, show that when $p\to\infty$ \begin{align}P\big(\max_{j=1,\ldots,p}\left|Z_j\right|\geq\sqrt{\alpha\log(p)}\big)\\&= 1 - \bigg(1 - P(\left|Z_1\right|\geq \sqrt{\alpha\log(p)}\big)\bigg)^p\\&=1 - \exp\bigg(-\sqrt{\dfrac{2}{\alpha\pi}}\dfrac{p^{1-\alpha/2}}{\sqrt{\log p}} + \mathcal{O}\bigg(\dfrac{p^{1-\alpha/2}}{(\log p )^{3/2}}\bigg)\bigg).\end{align}
I think I'm going in the right direction. This is what I've done so far:
\begin{align}P\big(\max_{j=1,\ldots,p}\left|Z_j\right|\geq\sqrt{\alpha\log(p)}\big) =& \,P(\left|Z_1\right|\geq\sqrt{\alpha\log(p)}, \ldots, \left|Z_n\right|\geq\sqrt{\alpha\log(p)}\big)\\ =&\,\bigg(1 - P(\left|Z_1\right|\leq \sqrt{\alpha\log(p)}\big)\bigg)^p\\=&\,1 - \bigg(1 - P(\left|Z_1\right|\geq \sqrt{\alpha\log(p)}\big)\bigg)^p \end{align} Furthermore, since the normal distribution is symmetrical we have that $P(\left|Z_i\right|\geq z) = 2P(Z_i\geq z)$. Hence, I arrive at $$1 - \bigg(1 - P(\left|Z_1\right|\geq \sqrt{\alpha\log(p)}\big)\bigg)^p = 1 - \bigg(1 - 2P(Z_1\geq \sqrt{\alpha\log(p)}\big)\bigg)^p$$ Unfortunately, I don't know how to proceed from here. I have no idea how to get to the exponential or arrive at the big $\mathcal{O}$ notation.
Question: How do I solve this exercise?