It's probably just a typo.
this question can be solved easily (by taking the derivative of the binomial theorem formula)
and multiplying both sides by $x$ you get
$nx(1+x)^{n-1} = \sum\limits_{k=1}^n k x^k{n\choose k}$.
If what is typed were also true we'd have
$ \sum\limits_{k=1}^n k x^k{n\choose k} = \sum\limits_{k=1}^n k x^k{x\choose k}$
.... for every possible value of $n$ and $x$. That's surely nuts!
A simple counter-example would be $x=1$.
The left hand side $n(1+1)^{n-1} = \sum\limits_{k=1}^n k {n\choose k}= n2^{n-1}$ for all $n$. (Mildly interesting... so $5*2^4 = 80$ is equal to ${5\choose 1}+2{5\choose 2} + 3{5\choose 3}+4{5\choose 4}+5{5\choose 5}=5+20 + 30 + 20+5$.... cool.... I guess.....That's a funny side shift I never noticed... well, I'll play with that later... but I digress)
But the right hand side is $\sum\limits_{k=1}^n k x^k{x\choose k}= 1*1^1{1\choose k} + \sum_{k=2}^n k1^k{1\choose k} = 1 + \sum_{k=1}^n 0 = 1$.
Obviously $n2^{n-1} =1$ is not a universal identity for all $n$ so the statement is false.
It's probably just a typo.