$\exp(2)$ does not converge $2$-adically. We have $\exp(2)= \sum_{i=0}^n {\frac{2^n}{n!}}$.
I am trying to show that $\exp(2)$ does not converge $2$-adically. 
i.e. I need to show $\nu_2 (\frac{2^n}{n!})$ does not tend to $\infty$ as $n\to \infty$, where $\nu_p (x) = \max$ { $a : p^a $divides $x$ }.
$\nu_p (x)$ is $\infty$ only when $x$ is $0$. 
However, since $\lim_{n \to \infty} \frac{2^n}{n!}=0$, I'll have $\nu_2 (\frac{2^n}{n!})\to \infty$ and hence $\exp(2)$ does converge $2$-adically, which is the exact opposite of what I'm trying to prove. 
What went wrong there?

$\nu_{2}(\frac{2^{n}}{n!})$ does not go to infinity as n goes to infinity.  The function is unbounded, true, but it is not monotonically increasing with $n$.  It drops when $n$ hits a power of $2$ and gives you a lot of factors of $2$ in the factorial.  In fact $\nu_{2}(\frac{2^{n}}{n!})$ drops all the way down to $1$ when $n$ is a power of $2$.
The exponential function converges in $2$-adics when the argument is a multiple of $4$.
