How prove this $x_{2^n}\ge 2^n-n+1$ the sequence $ (x_n)_{n\ge 1}$, $ x_n$ being
the exponent of 2 in the decomposition of the numerator of
$$\dfrac{2}{1}+\dfrac{2^2}{2}+\cdots+\dfrac{2^n}{n}$$
 goes to infinity as $ n\to\infty$.
Even more, prove that $$x_{2^n}\ge 2^n-n+1$$
My idea: maybe
$$\dfrac{2}{1}+\dfrac{2^2}{2}+\cdots+\dfrac{2^n}{n}=\dfrac{2^n}{n}\cdot M'?$$
where $M'$ is postive numbers.
so we after we replacing $n$ by $2^n$, then
$$\dfrac{2}{1}+\cdots+\dfrac{2^{2^n}}{2^n}=\dfrac{2^{2^n}}{2^n}M''=2^{2^n-n}M''$$
then I can't,Thank you for your help

This is a partial answer, just underlying an interesting technique.
$$\frac{a_n}{b_n}=\sum_{k=1}^n\frac{2^k}{k},\qquad (a_n,b_n)=1$$
is the opposite of the sum of the reciprocal of the roots of
$$P_n(x) = \prod_{k=1}^{n}\left(x+\frac{k}{2^k}\right)=2^{-\frac{n(n+1)}{2}}\prod_{k=1}^{n}(k+2^kx),$$
so
$$\frac{a_n}{b_n}=\frac{P_n'(0)}{P_n(0)}=\left.\frac{d}{dx}\left(\log P_n(x)\right)\right|_{x=0},\tag{1}$$
and we just need to estimate the $2$-adic height of $P_n(0)$ and $P_n'(0)$. The first task is quite easy:
$$\nu_2(P_n(0))=-\frac{n(n+1)}{2}+\sum_{j=1}^{\infty}\left\lfloor\frac{n}{2^j}\right\rfloor\leq -\frac{n(n-1)}{2},\tag{2}$$
while for the second one some insight is needed. We need a lower bound for 
$\nu_2(Q_n'(0))-\nu_2(n!)$, where:
$$ Q_n(x)=\prod_{k=1}^{n}(k+2^k x),$$
$$ Q_{n+1}'(0) = (n+1)\cdot Q_n'(0) + 2^{n+1}n!,$$
$$ \nu_2(Q_{n+1}'(0)) \geq \min\left(\nu_2(n+1)+\nu_2(Q_n'(0)),n+1+\sum_{j=1}^{\infty}\left\lfloor\frac{n}{2^j}\right\rfloor\right).$$
The last line trivially gives that the sequence $\delta_n=\nu_2(Q_n'(0))-\nu_2(n!)$ is non-decreasing, so, in order to prove the first point, we just need to prove the second one, that may be related to the structure of $\mathbb{Z}_{/2^n\mathbb{Z}}^*$.
