Stirling numbers of the second kind identity

I am understanding this a little better. Could someone let me know if I am on the right track with the following proof?

Give combinatorial proof: n$\displaystyle \geq$1 and k$\displaystyle \geq$1 then

$\displaystyle S(n,k)=\sum_{i=0}^{n-1}\left({n-1\atop i}\right)S(i,k-1)$.

Condition on whether the element n is in a block by itself. If it is, then all such partitions can be constructed by first specifying a (k-1) partition of [n-1] and then adding the block {n}. There are S(n-1,k-1) such partitions.

We can continue the next case by putting n and n-1 into a block by themselves. We construct these partitions by first specifying a (k-1) partition of [n-2] and then adding the block {n,n-1}. There are $\displaystyle \left({n-1\atop n-2}\right)S(n-2,k-1)$ such partitions. We can continue in this same manner until we reach a (k-1) partition of [n-n] by noting that S(n,k)=0 for all n<k. Then we use the sum principle for all terms.

Please note that in my text [n]={1,2,3,...,n}