Proving Pascal's identity $ \binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1}$ I'm trying to prove Pascal's identity, no luck so far. 
I have an answer which seem to include some unexplained shifts that I don't get.
What needs to be proved:
$$ \binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1}$$
This shift is not clear to me:
$$\binom{n}{k}+\binom{n}{k+1}=\frac{n!}{k!(n-k)!}+\frac{n!}{(k+1)!(n-k-1)!}$$
$$=\bbox[10pt,border:5px solid blue]{\frac{n!}{k!(n+1-(k+1))!}+\frac{n!(n-k)}{(k+1)!(n+1-(k+1))!}}$$
 A: Rather than using the factorial definition within this argument, consider a more combinatorial argument, perhaps. 
Given a set $S$ of order $n+1$, $\binom{n+1}{k+1}$ is the number of subsets of $S$ of order $k+1$. Fix one element of the set, $a \in S$. We have two disjoint and exhaustive possibilities here:


*

*We include $a$ in our subset, and choose the remaining $k$ from $S\backslash\{a\}$, giving us $\binom{n}{k}$ choices. 

*We do not include $a$, and choose all $k+1$ elements from the subset $S\backslash\{a\}$, giving us $\binom{n}{k+1}$ choices. 


Since these choices are disjoint and exhaustive, we can apply the elementary counting principle, the addition rule (aka the rule of sum), and the result follows.
A: Answering Eran's question:

The first term:
$\frac{n!}{k!(n-k)!} = \frac{n!}{k!(n+1-k-1)!} = \frac{n!}{k!(n+1-(k+1))!}$
is just adding and subtracting an extra $1$ and re-writing.

The second term:
$\frac{n!}{(k+1)!(n-k-1)!} = \frac{n!(n-k)}{k!(n-k-1)!(n-k)} = \frac{n!(n-k)}{k!(n-k)!}$
is multiplying $(n-k)$ in the numerator and denominator which allows you to re-write the factorial in the denominator. There's one last step, but that last step is the same as the first term.
A: Two facts seem to be involved in getting to the formula in the blue box from the previous formula. One fact is that 
$$n+1-(k+1)=n-k.\tag1$$
The other fact is that
$$ x((x-1)!) = x!.\tag2$$
I honestly don’t see why the fact in $(1)$ was invoked at this point. Yes, it’s a true fact, but invoking it just means we write $n+1-(k+1)$
in several places where $n-k$ would work better. Eventually, it might be useful to rewrite $n-k$ that way in order to relate to the expansion of $\binom{n+1}{k+1}$ into factorials, but we only actually make use of that in the last step of the derivation. 
In particular, the specific way that the fact in $(2)$ is invoked is that we can multiply 
$\frac{n!}{(k+1)!(n-k-1)!}$
by $n-k$ on the top and bottom. 
On the top we get $n!(n-k),$
as shown in the blue box, and on the bottom we get
$$(n-k)((n-k-1)!) = (n-k)!,$$
which uses fact $(2)$ with $x$ replaced by $n-k.$
Of course, because of fact $(1)$ It is equally true that 
$$(n-k)((n-k-1)!) = (n+1-(k+1))!,$$
which is how we get the formula in the blue box, but I think this is harder to see.
A: The term in the blue rectangle should be:
$$
\frac{n!(k+1)}{(k+1)!(n-k)!}+\frac{n!(n-k)}{(k+1)!(n-k)!}
$$
Now, it should be easy to add both fractions and arrive at the desired result. 
A: Considering the highest factors in the factorials,
$$\binom n{k+1}=\binom nk\frac{n-k}{k+1}$$
so that
$$\binom nk+\binom n{k+1}=\binom nk\frac{n+1}{k+1}=\binom{n+1}{k+1}.$$
