Pine tree shaped in binomial coefficients and a proving the formula derived from the shape

We have

$$\sum_{m=0}^{n-k} {n-m-1\choose k-1} = \sum_{m=0}^{n-k} {n-m-1\choose n-k-m} \\ = [z^{n-k}] (1+z)^{n-1} \sum_{m=0}^{n-k} z^m (1+z)^{-m} $$

Note that with the second form we have for $m$ in the range $[0,n-k]$ that $n-k-m$ is non-negative even when $n=k$ or $k=0.$ The coefficient extractor enforces the range and we may continue with

$$[z^{n-k}] (1+z)^{n-1} \sum_{m\ge 0} z^m (1+z)^{-m} \\ = [z^{n-k}] (1+z)^{n-1} \frac{1}{1-z/(1+z)} \\ = [z^{n-k}] (1+z)^{n-1} \frac{1+z}{1+z-z} \\ = [z^{n-k}] (1+z)^{n} = {n\choose n-k} = {n\choose k}$$

as claimed.

Remark. When $k=0$ we get $\sum_{m=0}^n \frac{(n-m-1)^\underline{n-m}}{(n-m)!}$, the terms of which are all zero except for $m=n$ which is $\frac{(-1)^{\underline{0}}}{0!} = 1$ for a total of one, which is in turn ${n\choose 0}$.