Proof for $\sum_{k=1}^n k2^k \binom{2n-k-1}{n-1} = n\binom{2n}{n}$?

We prose to solve this sum by making use of two results $$\sum_{p=m}^{2m} 2^{-p} {p \choose m}=1~~~~(1)$$ see How to prove that $\sum_{i=0}^n 2^i\binom{2n-i}{n} = 4^n$.

and

see Help in summing: $\sum_{k=m} ^{2m} ~k\cdot ~2^{-k} {k \choose m}$ $$\sum_{p=m}^{2m} p ~2^{-p} {p \choose m}=(2m+1)-2^{-2m-1}(m+1) {2m+2 \choose m+1}~~~~(2)$$

$$S=\sum_{k=1}^{n} k 2 ^k {2n-k-1\choose n-1}= 2^{2m+1} \sum_{p=m}^{2m}[2m+1-p]~2^{-p}~ {p \choose m},~ m=n-1, p=2n-k-1.~~~~~(3)$$ Next by using ((1) and (2) in (3) we get $$S=(m+1) {2m+2 \choose m+1}=n {2n \choose n}.$$

We seek to show that

$$\sum_{k=1}^n {2n-k-1\choose n-1} k 2^k = n {2n\choose n}.$$

The LHS is

$$\sum_{k=1}^n {2n-k-1\choose n-k} k 2^k \\ = [z^n] (1+z)^{2n-1} \sum_{k=1}^n k 2^k z^k (1+z)^{-k}.$$

Now we may extend $k$ to infinity because of the coefficient extractor in front:

$$[z^n] (1+z)^{2n-1} \sum_{k\ge 1} k 2^k z^k (1+z)^{-k} \\ = [z^n] (1+z)^{2n-1} \frac{2z/(1+z)}{(1-2z/(1+z))^2} \\ = [z^n] (1+z)^{2n} \frac{2z}{(1-z)^2} = 2\sum_{k=0}^n {2n\choose k} (n-k) \\ = 2n \sum_{k=0}^n {2n\choose k} - 2\sum_{k=1}^n {2n\choose k} k = 2n \left(\frac{1}{2} 2^{2n}+\frac{1}{2} {2n\choose n}\right) - 4n\sum_{k=1}^n {2n-1\choose k-1} \\ = n 2^{2n} + n {2n\choose n} - 4n\sum_{k=0}^{n-1} {2n-1\choose k} = n 2^{2n} + n {2n\choose n} - 4n \frac{1}{2} 2^{2n-1} \\ = n {2n\choose n}.$$

This is the claim.