To prove the binomial identity $$\sum_{m=k}^{n-1}\binom{m}{k} = \binom{n}{k+1}$$ will an inductive method on $n-1$ be valid?
Specifically, if we prove the base case where $n-1 = 0$, to determine it will hold for $0\leq k\leq0$
$$\sum_{m=k}^0\binom{m}{k} = \begin{cases} 1 &\mbox{if } k=0 \\ 0 &\mbox{if }k>0 \end{cases}$$
and we know that on the right-hand side $\binom{1}{k}$ will produce the same results case-wise.
Then we take the hypothesis that it holds for $0\leq k \leq n-2$, where $n-2 >0$ $$\sum_{m=k}^{n-2}\binom{m}{k} = \binom{n-1}{k+1}$$
And finally if we use this assumption to finish the inductive step, proving it holds for $0\leq k\leq n-1$ $$\sum_{m=k}^{n-1}\binom{m}{k} = \sum_{m=k}^{n-2}\binom{m}{k} + \binom{n-1}{k}$$ which by Pascal's rule gives the final result $$\binom{n-1}{k+1} +\binom{n-1}{k} = \binom{n}{k+1}$$
Does this abide by the logic of induction?
We'll look at the case when $k<n$, since in all other cases both sides are zeroes.
I would recommend you to do the induction on $n$. Then if it holds for $n-1$ use induction that it holds for $n$, while $0\le k \le n-2$ and look at the case $k=n-1$ separately, since the inductive hypothesis doesn't cover that case. Here's a little help
Induction Base Let $n=1$, then we have $k=0$, which gives us:
$$\sum_{m=0}^0 \binom{m}{0} = \binom{1}{1}$$
Which is true since boths sides are 1.
Induction Hypothesis
Let the equality holds for $0\le n \le s-1$, while $0\le k < n$
Inductive Step
We'll know prove that it holds for $n=s$. We'll check 2 separate cases:
Case 1: $0\le k<s-1$
This makes sure that the inductive step is covered by the hypothesis. So now we have:
$$\sum_{m=k}^{s-1} \binom{m}{k} = \binom{s}{k+1}$$ $$\sum_{m=k}^{s-2} \binom{m}{k} + \binom{s-1}{k} = \binom{s}{k+1}$$
Now use the hypothesis and we have:
$$\binom{s-1}{k+1} + \binom{s-1}{k} = \binom{s}{k+1}$$
which hold from the Pascal's rule.
Case 2: $k=s-1$
The equality now looks like:
$$\sum_{m=s-1}^{s-1} \binom{m}{s-1} = \binom{s}{s-1+1}$$ $$\binom{s-1}{s-1} = \binom{s}{s} = 1$$
Hence the proof.
Note that you can use the fact that if $k\ge n$ both sides are zeroes and extend the equality to $k\in \mathbb{N}_0$ so you wouldn't have to check different cases, since they will all be covered by the hypothesis. But I wanted to give you a nice, more formal proof that you can use in other problems, since you may find an equality that doesn't hold for all fixed numbers in $\mathbb{N}_0$