Let $\displaystyle \xi_1,\xi_2,...$ be independent identically distributed Gaussian variables with mean zero and variance one. Let $\displaystyle \eta_1,\eta_2,...$ be independent identically distributed exponential random variables with mean one. How do you prove that there is $\displaystyle n>0$ such that:

$\displaystyle \mathbb{P}(max(\eta_1,...,\eta_n)\geq max(\xi_1,...,\xi_n))>0.99$

I thought a proper application of the Law of Total Probability would do the trick, but things don't seem that simple. Otherwise, some fancy convolution... But I'm probably wrong.

Thanks for your help.