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

Now,I have solution This nice  problem,
 we only note that
$$\dfrac{2}{1}+\dfrac{2^2}{2}+\cdots+\dfrac{2^n}{n}=\dfrac{2^n}{n}\sum_{k=0}^{n-1}
\dfrac{1}{\binom{n-1}{k}}$$
This indentity  proof can see:http://www.artofproblemsolving.com/Forum/viewtopic.php?p=371496&sid=e0319f030d85bf1390a8fb335fd87c9d#p371496
also I remember this indentity stackmath have post it,But I can't find this link.
