(Combinatorial) proof of an identity of McKay Lemma 2.1 of this paper claims that for integer $s>0$ and $v \in \mathbb{N}$, we have
$$
\sum_{k=1}^s \binom{2s-k}{s} \frac{k}{2s-k} v^k (v-1)^{s-k}
= v \sum_{k=0}^{s-1} \binom{2s}{k} \frac{s-k}{s} (v-1)^k
$$
The author gives a combinatorial interpretation of the left hand side in terms of closed walks on a class of graphs that are locally acyclic in a particular sense, but omits the proof of equality. I've hit a wall trying to prove this, so I was wondering what ideas others have. Combinatorial proofs welcomed!
 A: Suppose we seek to verify that
$$\sum_{k=1}^n {2n-k\choose n} \frac{k}{2n-k} v^k (v-1)^{n-k}
= v \sum_{k=0}^{n-1} {2n\choose k} \frac{n-k}{n} (v-1)^k.$$
Now the LHS is
$$\frac{1}{n}\sum_{k=1}^n {2n-k-1\choose n-1} k v^k (v-1)^{n-k}.$$
Re-write this as
$$\frac{1}{n}\sum_{k=1}^n {2n-k-1\choose n-k} k v^k (v-1)^{n-k}.$$
Introduce 
$${2n-k-1\choose n-k} = 
\frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{w^{n-k+1}} (1+w)^{2n-k-1} \; dw.$$
Observe that this zero when $k\gt  n$ so we may extend $k$ to infinity
to obtain for the sum
$$\frac{1}{n} \frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{w^{n+1}} (1+w)^{2n-1} 
\sum_{k\ge 1} k v^k (v-1)^{n-k} \frac{w^k}{(1+w)^k}\; dw
\\ = \frac{(v-1)^n}{n} \frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{w^{n+1}} (1+w)^{2n-1} 
\frac{vw/(v-1)/(1+w)}{(1-vw/(v-1)/(1+w))^2} \; dw
\\ = \frac{(v-1)^n}{n} \frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{w^{n+1}} (1+w)^{2n} 
\frac{vw(v-1)}{((v-1)(1+w)-vw)^2} \; dw
\\ = v\frac{(v-1)^{n+1}}{n} \frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{w^{n}} (1+w)^{2n} 
\frac{1}{(-1-w+v)^2} \; dw
\\ = v\frac{(v-1)^{n-1}}{n} \frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{w^{n}} (1+w)^{2n} 
\frac{1}{(1-w/(v-1))^2} \; dw.$$
Extracting the coefficient we obtain
$$v\frac{(v-1)^{n-1}}{n} 
\sum_{q=0}^{n-1} {2n\choose q} 
\frac{(n-1-q+1)}{(v-1)^{n-1-q}}
\\ = v
\sum_{q=0}^{n-1} {2n\choose q} 
\frac{(n-1-q+1)}{n} (v-1)^q
\\ = v
\sum_{q=0}^{n-1} {2n\choose q} 
\frac{(n-q)}{n} (v-1)^q.$$
This concludes the argument.
