How to count partitioned subsets of a binomial coefficient I have been reading Dr. Carl Wagner's book Basic Combinatorics, and I cannot wrap my head around a particular theorem. Someone else has asked a very similar question and accepted an answer, but I'm afraid I still don't understand.
As stated, the theorem runs thus:
For all $ n, \; k \in \mathbb{N} $
$ \displaystyle \sum_{j = 0}^{n} \binom{j}{k} = \displaystyle \sum_{j = k}^{n} \binom{j}{k} = \binom{n + 1}{k + 1} $
There follows a proof of the $\binom{n + 1}{k + 1} $ bit using some handy substitutions. What confuses me is the next part:

#2. Clearly $ \binom{n+1}{k+1} $ counts the class of all (k + 1)-element subsets of [n + 1]. But this class of k+1 subsets may be partitioned into subclasses corresponding to j = k, k + 1, . . . , n as follows.
The subclass of subsets with largest element equal to k + 1 is counted by $ \binom{k}{k} $ (why?), the subclass of subsets with largest element equal to k + 2 is counted by $ \binom{k+1}{k} $ (why?), . . ., and the subclass of subsets with largest element equal to n + 1 is counted by $ \binom{n}{k} $ (why?).

I don't understand why at all! Indeed, it makes no sense to me.
It seems to me that the number of subsets of [n+1] with largest element k+1 would be $\binom{k+1}{k+1}$. Sure, that's the same as $ \binom{k}{k} $  since they both equal one, but that just seems to be a coincidence. We're choosing k+1 numbers from a set that runs from 1 to k+1. There can, naturally, be only one set that fits that bill. Yet to use $ \binom{k}{k} $ to represent that one set seems confusing to me, and every example that follows makes less sense.
What am I missing?
 A: The largest element of a set has to be in the set.  If you specify that $j+1$ is the largest element of a $(k+1)$-subset, it means that you have only $j$ choices $\{1,\dots,j\}$ for the remaining $k$ elements.  The three examples given correspond to $j=k$, $j=k+1$, and $j=n$.
A: Let's say $n = 10$ and you want a subset of $4$ elements, the largest of which you want to be $6$.
Now that is not the same thing as choosing a subset of $\{1,2,3,4,5,6\}$ with $4$ elements because that would not guarantee that the largest element is $6$, only that the largest element is $6$ or less.
Here are all the $4$ element subsets whose largest element is $6$:
$$1236, 1246, 1256, 1346, 1356, 1456, 2346, 2356, 2456, 3456$$
Additionally, here are the $4$ element subsets of $\{1,2,3,4,5,6\}$ which were not listed already:
$$1234, 1235, 1245, 1345, 2345.$$
The important observation is that if you want $6$ to be in your set and you have no choice about the matter, then you are not choosing $6$. You choose $3$ elements of $\{1,2,3,4,5\}$ and then you get an extra $6$ whether you like it or not. And you will agree that we can pair everything in that first list with:
$$123, 124, 125, 134, 135, 145, 234, 235, 245, 345$$
simply by adding or removing the $6$ that is required to be there (not chosen).
