Prove that $\liminf \frac{M_n}{\sqrt{2\log n}}\geq 1$, where $M_n=\max_{1}^n X_i$ and $(X_i)$ is a sequence of i.i.d standard normal random variables Question

Let $(X_n)_{n\geq 1}$ be an i.i.d sequence of standard normals. Show that with probability one $\liminf \frac{M_n}{\sqrt{2\log n}}\geq 1$, where $M_n=\max_{1}^n X_i$.

My attempt
Given $\varepsilon >0$, it suffices to show that $P(\frac{M_n}{\sqrt{2\log n}}<1-\varepsilon \quad \text{i.o.})= 0$. To this end put $A_n=(\frac{M_n}{\sqrt{2\log n}}<1-\varepsilon)$ and we attempt to use the Borel-Cantelli Lemma. So
$$
\sum_1 ^\infty P(A_n)=\sum_{1}^\infty P(X_1<c_n)^n=\sum_{1}^\infty (1-\bar{\Phi} (c_n))^n
$$
where $c_n=(1-\varepsilon )(\sqrt{2\log n})$ and $\bar{\Phi}=1-\Phi$ is the survival function. At this point since the normal distribution has no nice form for its cdf I have to use some argument involving asymptotics of this sum. To this end, I know that
$$
\bar{\Phi}(x)\sim\frac{\phi (x)}{x} 
$$ 
as $x\to \infty$ where $\phi$ is the density of a standard normal. More precisely,
$$
\left(\frac{1}{x}-\frac{1}{x^3}\right)\leq \frac{\bar{\Phi}(x)}{\phi(x)}\leq \frac{1}{x}.\tag{1}
$$
We can write 
$$
\sum_{1}^\infty P(X_1<c_n)^n\leq \sum_{1}^\infty(1-(c_n^{-1}-c_n^{-3})\phi(c_n))^n
$$
using $1$, but I am not sure how to argue that this is finite.
 A: For any $x \geq 0$ it holds that
$$1-x \leq e^{-x}. \tag{1}$$
For $x:= \mathbb{P}(X_1 \geq c_n)$ this implies that
$$\mathbb{P}(X_1 < c_n)^n = (1-\mathbb{P}(X_1 \geq c_n))^n \stackrel{(1)}{\leq} \exp \left(-n \mathbb{P}(X_1 \geq c_n) \right). \tag{2}$$ As
$$\begin{align*} \mathbb{P}(X_1 \geq c_n) &= \frac{1}{\sqrt{2\pi}} \int_{c_n}^{\infty} \exp \left( - \frac{y^2}{2} \right) \, dy \\ &\geq \frac{1}{\sqrt{2\pi}} \frac{c_n}{c_n^2+1} \exp \left(- \frac{c_n^2}{2} \right) \\ &\geq \frac{1}{2\sqrt{2\pi}} \frac{1}{(1-\epsilon) \sqrt{2 \log n}} \frac{1}{n^{(1-\epsilon)^2}} \end{align*}$$
we have
$$n \mathbb{P}(X_1 \geq c_n) \geq \frac{1}{2\sqrt{2\pi}}  \frac{1}{(1-\epsilon) \sqrt{2 \log n}} n^{\epsilon}$$
for $\epsilon \in (0,1)$. For $n \geq N=N(\epsilon)$ sufficiently large this shows that
$$n \mathbb{P}(X_1 \geq c_n) \geq C n^{\epsilon/2}$$
where $C=C(\epsilon) := 1/(2 \sqrt{4\pi} (1-\epsilon))$. Plugging this estimate into $(2)$ we find that
$$\sum_{n \geq N} \mathbb{P}(X_1>c_n)^n \leq \sum_{n \geq N} \exp(- C n^{\epsilon/2})< \infty.$$
