Maximum of exponential random variables is bigger than maximum of normal random variables almost surely as $n$ tends to infinity. 
Let $X_1,X_2,\cdots,X_n$ be iid Gaussian random variables with mean $0$ and variance $1$. Let $Y_1,Y_2,\cdots,Y_n$ be iid exponential random variables with mean $1$. Prove that $$\lim_{n\to\infty}P\left(\max{(Y_1,Y_2,\cdots,Y_n)}\ge \max{(X_1,X_2,\cdots,X_n)}\right)=1$$

I let $W_n=\max{(X_1,X_2,\cdots,X_n)}$ and $Z_n=\max{(Y_1,Y_2,\cdots,Y_n)}$.As $X_i$'s and $Y_i$'s are iid, it is easy to get the distribution of $W$ and $Z$ as $F_W(w)=\left(F_X(w)\right)^n$ and $F_Z(z)=\left(F_Y(z)\right)^n$. I computed $$P(Z_n\ge W_n)=\int_{-\infty}^{\infty}F_W(w)f_Z(w)dw=\int_0^{\infty}\left(\Phi(w)\right)^nn(1-e^{-w})^{n-1}e^{-w}dw$$ where $\Phi$ is the CDF of standard normal distribution. I am not sure how to estimate the resulting integral as $n\to\infty$, although I could verify that it converges to 1 by using numerical computation.
Also, I believe that there should be a more clever and elegant way to deal with this problem, but I don't know how. Thanks for any hint in advance.
 A: The question does not explain if $(X_n)$ should be assumed independent of $(Y_n)$, and in fact this hypothesis is not necessary for the desired result to hold. 
To see why, note that for every $x$, $$P(\max(Y_1,\ldots,Y_n)<\max(X_1,\ldots,X_n))\leqslant P(\max(X_1,\ldots,X_n)>x)+P(\max(Y_1,\ldots,Y_n)<x)$$ with $$P(\max(X_1,\ldots,X_n)>x)=1-(1-P(X_1>x))^n$$ and $$P(\max(Y_1,\ldots,Y_n)<x)=P(Y_1<x)^n$$ Thus, if one can choose some sequence $(x_n)$ such that the convergences $$(1-P(X_1>x_n))^n\to1\qquad\&\qquad P(Y_1<x_n)^n\to0$$ hold simultaneously, we are done. To do so, recall that, for every $x>1$, $$P(X_1>x)\leqslant e^{-x^2/2}\qquad\&\qquad P(Y_1>x)=e^{-x}$$ thus, $$(1-P(X_1>x_n))^n\geqslant(1-e^{-x_n^2/2})^n\qquad\&\qquad P(Y_1<x_n)^n=(1-e^{-x_n})^n$$ hence the two convergences we need are simultaneously realized as soon as $x_n\to\infty$ with $$ne^{-x_n^2/2}\to0\qquad\&\qquad ne^{-x_n}\to\infty$$ that is, $$e^{x_n}\ll n\ll e^{x_n^2/2}$$ or, equivalently, $$x_n^2-2\log n\to\infty\qquad\&\qquad x_n-\log n\to-\infty$$ which happens, say, if $$x_n\sim\tfrac12\log n$$ Thus, as desired, $$P(\max(Y_1,\ldots,Y_n)<\max(X_1,\ldots,X_n))\to0$$
