For the RHS -Sorry; I will give you a little more detail.
So I have the definition
[MATH] \binom{n + 1}{r} = \binom{n}{r} + \binom{n}{r-1} [/MATH]
and I gathered that I could manipulate one side of the equation to make it equal to the other that would be my proof by induction.
So it tried manipulating the right side like so:
[MATH] \frac{(n+1)!}{r!(n+1-r)!} + \frac{(n+1)!}{(r-1)!(n-(r-1))!} [/MATH]
and worked with that for about half an hour but couldn't make the two sides of the equations equate.
I figure I am either just not approaching the question correctly or there is some element of factorial manipulation that I am unaware of.
I probably should have recognized from your use of "Pascal" in the title that you are starting fro Pascal's triangle itself as the definition; that implies other recursion formulas besides the one quoted, but we can infer that you are expected to use that one (together with other information) for your proof.
Did you complete the proof, or do you need additional help? If so, we'll want to see what you've done again.
Rather than changing the numerators first, how about first making a common denominator, so you can add the fractions and get the single fraction on the LHS that you are aiming for? Things will be a little simpler that way.
You'll be using facts like r! = r(r-1)! .
You haven't actually shown the induction process, but I assume this step is part of a fuller answer you are working on.