Proving Pascal's identity

To include more intuition. Let's say you want to know number of ways you can pick $k$ objects out of $n$ objects set. Assume you are holding one object. Either you want this object to be in your subset, or you want it excluded.

If you want this object to be in your subset, then there are $\binom{n-1}{k-1}$ remaining choices. Why? You include the one you were holding, so now you have $n-1$ left in the set. And since you decided to include it, you only have to pick $k-1$ more.

If you do not want this object to be in your subset, then there are $\binom{n-1}{k}$ remaining choices. Why? You are holding one object, so there are $n-1$ remaining, and you don't want it to be included in your subset, so you still have to pick $k$ objects from that $n-1$ set.

And both cases above represent "$n$ choose $k$" number of ways to pick $k$ objects from the set of $n$ objects. Hence the identity.

The algebra isn't really cumbersome at all

$$\binom{n}{r} + \binom{n}{r-1} = \frac{n!}{r!(n-r)!} + \frac{n!}{(r-1)!(n-r+1)!} \\ = \frac{n!(n-r+1) + n!(r)}{r!(n-r+1)!} = \frac{(n+1)!}{r!(n+1-r)!} = \binom{n+1}{r} $$

$(T+1)^{n+1}=(T+1) (T+1)^n$ and binomial coefficient are...