Asymptotic probability concerning the largest absolute value in an iid Gaussian sample Let $v[n]$ be a vector $N$ $iid$ gaussian samples, of ~$N(0,\sigma^2)$ Also, let $v_{max}$ denote the maximum absolute value of all the samples given. (That is, if I took the absolute value of all the $N$ random samples given, and picked the max, that would be $v_{max}$).
I am trying to ascertain why this statement is true:
In English:
"The probability of $v_{max}$ exceeding $\sigma \sqrt{2log_e N}$ approaches $0$, as the sample size $N$ approaches infinity."
In mathematical terms:
$$
\lim_{N \to \infty} P \left( v_{max} > \sigma \sqrt{2log_e{N}} \right) = 0
$$
I am not even sure where to start on this. Intuitively, I am not even sure why the sample size $N$ should matter, so I do not even have a starting intuitive traction for this particular problem. Would appreciate any insights. 
TLDR: How is the $\sigma \sqrt{2log_e{N}}$ derived exactly?

Edit, what I have tried so far:
Thanks to the comments, I have managed to piece together something, but it still seems I am a ways away.
I want to show that $\lim_{N \to \infty} P(v_{max} > z) = 0$ becomes true when $z=\sigma \sqrt{2log_eN})$. Based on the $iid$ assumption, we can say then say that 
$$P(v_{max} > z) = P(|v_1|, |v_2|, |v_3| ... |v_N| > z) = \left [2 - 2\Phi_{\sigma}(z) \right]^N$$
, where $\Phi_{\sigma}$ is the CDF the common gaussian distribution, given by:
$$
\Phi_{\sigma}(z) = \frac{1}{2} + \frac{1}{2} erf \left [\frac{z}{\sigma \sqrt{2}} \right]
$$
Then this means, that by simple substitution,
$$
P(v_{max} > z) = \left[1 - erf \left( \frac{z}{\sigma \sqrt{2}}\right) \right]^N
$$
And thus, 
$$
\lim_{N \to \infty} P(v_{max} > z) = \lim_{N \to \infty} \left[1 - erf \left( \frac{z}{\sigma \sqrt{2}}\right) \right]^N = 0
$$
Here however I am stuck. How does one show from here that $z=\sigma \sqrt{2log_eN})$?...
Thanks
 A: Expanding on whuber's comment and my minor emendation of it, we have that for $z \geq 0$,
$$P\{v_\max < z\} = [2\Phi(z/\sigma)-1]^N = [1-2Q(z/\sigma)]^N$$ where
$\Phi(x)$ is the standard normal distribution function and $Q(x)=1-\Phi(x)$ is
its complement. Now, since $1-2Q(z) < 1$ for $z > 0$, if $z$ and $\sigma$
were  constants, that is, not functions of $N$, then as we raised a quantity
less than $1$ to ever higher powers $N$, the result would tend to $0$ as $N \to \infty$.
On the other hand, if $z/\sigma$ were such that $2Q(z/\sigma)=aN^{-1}$, then we
could use the standard result
$$\lim_{N\to \infty}\left(1-\frac{a}{N}\right)^N = e^{-a}$$
to deduce that $P\{v_\max < z\}$ converges to a positive constant (smaller
than $1$) as $N \to \infty$ and so  $P\{v_\max > z\}$ converges to a constant
greater than $0$. This is not quite good enough; we need to have 
$2Q(z/\sigma)$ go to $0$ slightly
faster than as $N^{-1}$.  What will work?  Well, a standard but weak
bound on $Q(x)$ is $$Q(x) < \frac{1}{2}e^{-x^2/2} ~~ \text{for} ~x > 0$$
and so if we choose $z = \sigma\sqrt{2\ln N}$, we get 
$2Q(z/\sigma) < e^{-\ln N} = N^{-1}$ which we just said is not a fast
enough decay with increasing $N$, but if we use the
tighter bound
$$Q(x) < \frac{1}{x\sqrt{2\pi}}e^{-x^2/2},$$
with the same choice $z = \sigma\sqrt{2\ln N}$,
that extra $x$ in the denominator is giving us
an $\sqrt{\ln N}$ faster decay that suffices to allow us
to conclude that
$$P\{v_\max > \sigma\sqrt{2\ln N}\,\} \to 0~~ \text{as}~ N \to \infty.$$
