I would like to prove the following equality: $$\sum_k (-1)^{n-1}(-2)^k\binom{n}{k+1}\binom{n+k-1}{k}=\sum_{k=0}^{n-2}\binom{2n-k-2}{n-1}\binom{n-2}{k}$$
but the power over two and the switch on the number of sums bothers me. Any help would be welcome.
(the equality it part of Note 1.41 in the book "Analytic Combinatorics")
The following answer is purely algebraic. We transform both sides of OPs expression to finally obtain the same representation. We also use the coefficient of operator $[z^n]$ to denote the coefficient $a_n$ of $z^n$ of a series $\sum_{k=0}^{\infty}a_kz^k$.
At first we transform the right-hand side. It's the easier one.
\begin{align*} \sum_{k=0}^{n-2}&\binom{2n-k-2}{n-1}\binom{n-2}{k}\\ &=\sum_{k=0}^{n-2}\binom{n+k}{n-1}\binom{n-2}{k}\tag{1}\\ &=\sum_{k=0}^{n-2}[z^{n-1}](1+z)^{n+k}\binom{n-2}{k}\tag{2}\\ &=[z^{n-1}](1+z)^n\sum_{k=0}^{n-2}\binom{n-2}{k}(1+z)^{k}\\ &=[z^{n-1}](1+z)^n(2+z)^{n-2}\tag{3}\\ &=\sum_{k=0}^{n-1}\left([z^k](1+z)^n\right)\left([z^{n-1-k}](2+z)^{n-2}\right)\\ &=\sum_{k=0}^{n-1}\binom{n}{k}\binom{n-2}{n-1-k}2^{(n-2)-(n-1-k)}\\ &=\sum_{k=1}^{n-1}\binom{n}{k}\binom{n-2}{k-1}2^{k-1}\tag{4}\\ \end{align*}
Comment:
In (1) we change the order of summation by transforming the index $k$ with $n-2-k$
In (2) we represent the binomial coefficient $\binom{n+k}{n-1}$ as the coefficient of $z^{n-1}$ of $(1+z)^{n+k}$
In (3) we write the sum as polynomial $(2+z)^{n-2}$.
And here's the left-hand side.
\begin{align*} \sum_{k=0}^{n-1}&(-1)^{n-1}(-2)^k\binom{n}{k+1}\binom{n+k-1}{k}\\ &=(-1)^{n-1} \sum_{k=0}^{n-1}(-2)^k\binom{n}{k+1}\binom{n+k-1}{n-1}\\ &=(-1)^{n-1} \sum_{k=1}^{n}(-2)^{k-1}\binom{n}{k}\binom{n+k-2}{n-1}\\ &=\frac{1}{2}(-1)^{n-1} \sum_{k=1}^{n}(-2)^{k}\binom{n}{k}[z^{n-1}](1+z)^{n+k-2}\\ &=\frac{1}{2}(-1)^{n-1} [z^{n-1}](1+z)^{n-2}\sum_{k=1}^{n}\binom{n}{k}(-2)^k(1+z)^{k}\\ &=\frac{1}{2}(-1)^{n-1} [z^{n-1}](1+z)^{n-2}\left\{(-1-2z)^n-1\right\}\\ &=\frac{1}{2} [z^{n-1}](1+z)^{n-2}\left\{(1+2z)^n+1\right\}\tag{5}\\ &=\frac{1}{2} [z^{n-1}](1+z)^{n-2}(1+2z)^n\\ &=\frac{1}{2}\sum_{k=0}^{n-1}\left([z^k](1+2z)^n\right)\left([z^{n-1-k}](1+z)^{n-2}\right)\\ &=\frac{1}{2}\sum_{k=0}^{n-1}\binom{n}{k}2^k\binom{n-2}{k-1}\tag{6}\\ \end{align*}
Since the expressions (4) and (6) are equal, the claim follows.
The summand with $k=0$ does not contribute anything. So we can start with $k=1$ and after an index transformation we obtain the somewhat more convenient representation \begin{align*} \sum_{k=0}^{n-2}\binom{n}{k+1}\binom{n-2}{k}2^k \end{align*}
Comment:
It seems there are some typesetting errors in this equality.
Suppose we seek to verify the closely related equality (obtained by a shift by one on the right)
$$(-1)^{n-1} \sum_{k=0}^{n+1} {n+1\choose k} (-2)^k {n+k-1\choose k} = \sum_{k=0}^{n-1} {2n-k\choose n-1} {n-1\choose k}.$$
Introduce for the LHS $${n+k-1\choose k} = {n+k-1\choose n-1} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^n} (1+z)^{n+k-1} \; dz.$$
This yields for the sum $$(-1)^{n-1} \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^n} (1+z)^{n-1} \sum_{k=0}^{n+1} {n+1\choose k} (-2)^k (1+z)^k\; dz \\ = (-1)^{n-1} \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^n} (1+z)^{n-1} (-1-2z)^{n+1} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^n} (1+z)^{n-1} (1+2z)^{n+1} \; dz.$$
Introduce for the RHS $${2n-k\choose n-1} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^n} (1+z)^{2n-k} \; dz.$$
This yields for the sum $$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^n} (1+z)^{2n} \sum_{k=0}^{n-1} {n-1\choose k} \frac{1}{(1+z)^k} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^n} (1+z)^{2n} \left(1+\frac{1}{1+z}\right)^{n-1} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^n} (1+z)^{n+1} (2+z)^{n-1} \; dz.$$
Put $z=2w$ in this integral to obtain $$\frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{1}{2^n w^n} (1+2w)^{n+1} (2+2w)^{n-1} \; 2 dw \\ = \frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{1}{w^n} (1+2w)^{n+1} (1+w)^{n-1} \; dw.$$
This concludes the argument.
Addendum. Apparently there is another possible version of this equality which is
$$(-1)^{n-1} \sum_{k=0}^{n-1} {n\choose k+1} (-2)^k {n+k-1\choose k} = \sum_{k=0}^{n-2} {2n-k-2\choose n-1} {n-2\choose k}.$$
Re-write the LHS as $$(-1)^{n-1} \sum_{k=1}^{n} {n\choose k} (-2)^{k-1} {n+k-2\choose k-1}.$$
We may extend this to include zero because the second binomial coefficent is zero then, getting $$(-1)^{n-1} \sum_{k=0}^{n} {n\choose k} (-2)^{k-1} {n+k-2\choose k-1}.$$
Introduce for the LHS $${n+k-2\choose k-1} = {n+k-2\choose n-1} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^n} (1+z)^{n+k-2} \; dz.$$
This yields for the sum $$(-1)^{n-1} \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^n} (1+z)^{n-2} \sum_{k=0}^n {n\choose k} (-2)^{k-1} (1+z)^k \; dz \\ = \frac{1}{2} (-1)^{n} \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^n} (1+z)^{n-2} (-1-2z)^n \; dz \\ = \frac{1}{2} \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^n} (1+z)^{n-2} (1+2z)^n \; dz.$$
For the RHS we introduce $${2n-k-2\choose n-1} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^n} (1+z)^{2n-k-2} \; dz.$$
This yields for the sum $$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^n} (1+z)^{2n-2} \sum_{k=0}^{n-2} {n-2\choose k} \frac{1}{(1+z)^k} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^n} (1+z)^{2n-2} \left(1+ \frac{1}{1+z}\right)^{n-2} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^n} (1+z)^{n} (2+z)^{n-2}\; dz.$$
Put $z=2w$ in this integral to get $$\frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{1}{2^n w^n} (1+2w)^{n} (2+2w)^{n-2}\; 2dw \\ = \frac{1}{2} \frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{1}{w^n} (1+2w)^{n} (1+w)^{n-2}\; dw.$$
This once more concludes the argument.
