Of the identity ${n \choose k} = {\sum\limits_{m=k-1}^{n-1} {m \choose k-1}}$, a proof by induction over $d=n-k$ should be possible. I can only see how $${n+1 \choose k} = {n \choose k} + {n \choose k-1}$$ (by Pascal's identity)
$$ ={\sum_{m=k-1}^{n-1} {m \choose k-1}} + {n \choose k-1}$$ (by induction hypothesis) $$={\sum_{m=k-1}^{n} {m \choose k-1}}$$ leads to the result. Can someone point me into the right direction for an induction over $d=n-k$?
On
You ask for an inductive proof, but I cannot resist a combinatorial proof when the opportunity presents itself.
The lefthand side, $\binom{n}{k}$, counts the number of $k$-element subsets of the set $\{1, 2, \dots, n\}$. Partition the collection of all such subsets according to their largest element. I will demonstrate.
The number of $k$-element subsets whose largest element is $k$ is $\binom{k-1}{k-1}$; choose $k-1$ elements from the set $\{1, 2, \dots, k-1\}$ and then include the element $k$. (This can be done in only one way, of course.)
The number of $k$-element subsets whose largest element is $k+1$ is $\binom{k}{k-1}$; choose $k-1$ elements from the set $\{1, 2, \dots, k\}$ and then include the element $k+1$.
In general, the number of $k$-element subsets whose largest element is $m+1$ is $\binom{m}{k-1}$; choose $k-1$ elements from the set $\{1, 2, \dots, m\}$ and then include the element $m+1$. Summing over $m$ gives the desired identity.
Since $d=n-k$, you can rewrite it as
$$\binom{d+k}k=\sum_{m=k-1}^{d+k-1}\binom{m}{k-1}\;;$$
let $\ell=m-(k-1)$, and this can in turn be rewritten
$$\binom{d+k}k=\sum_{\ell=0}^d\binom{\ell+k-1}{k-1}\;,$$
and you want to prove this by induction on $d$. Just use the same approach that you took for your induction on $n$.