<s>[INST] Show $\sum_{k=0}^{n} \binom{x}{k} \binom{y}{n-k} = \binom{x+y}{n}$. For $n\ge 0$, show $\sum_{k=0}^{n}  \binom{x}{k}  \binom{y}{n-k} = \binom{x+y}{n}$.
Over a well ordered set of non-negative integers, induction can be used.
Base case: $n=0$
$\binom{x}{0}  \binom{y}{0} = 1.\frac{(y)!}{(y)! \,\,(0)!}=1$.
Taking the rhs, get :
$\binom{x+y}{0} = 1$.
Induction hypothesis: it is true for natural number $n$.
So, $\sum_{k=0}^{n}  \binom{x}{k} \binom{y}{n-k} = \binom{x+y}{n}$
Inductive step: need show for next natural $n+1$
$\sum_{k=0}^{n+1} \binom{x}{k} \binom{y}{n-k+1} = \binom{x+y}{n+1}$
This should lead to :
$\binom{x+y}{n} + \binom{x}{n+1}\binom{y}{0}= \binom{x+y}{n+1}$
get on lhs:
$\binom{x+y}{n} + 1.\binom{x}{n+1}$
$\implies \frac{(x+y)!}{n!(x+y-n)!} + \frac{x!}{(n+1)!(x-n-1)!}$
Please help to proceed further, or suggest alternative approach.
 [/INST] There are much better proof strategies in my opinion, but here's an inductive proof of the Vandermonde convolution identity.
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