A nice identity involving urns and balls problem Prove the identity: $$\frac{\displaystyle\sum_{k=0}^{a} {n+a-k-2\choose n-2}}{\displaystyle {n+a-1\choose a}} = 1$$ where $C_{i}^{j}$ is defined as the number of ways to simultaneously choose $j$ objects from $i$ objects.
My attempt: I was trying to use a combinatorial argument by saying that out of the given $n$ balls, each of the terms within the summation is the probability of a given urn containing exactly $k$ balls for $k = 0,1,2,...,n$. Thus, the sum reflects the sum of all the probabilities of that urn having exactly $k$ balls, so it must be $1$. 
My question: I would like to see an algebraic proof for this identity. Could someone please help with such a proof? In case my argument above is incorrect, please help point out the mistake.

${n+a-1 \choose n-1}={n+a-1 \choose a}$ is the number of non-negative integer solutions to:
$x_1+x_2+\ldots+x_n=a$
Let $x_1=k$ with $k \in \{0,1,\ldots,a\}$. Then the corresponding number of solutions is ${n+a-k-2 \choose n-2}$.
