Prove by induction that $(1 + x)^n = ...$ (binomial expansion)

1. First, let $\binom{n}{k} = \frac{n!}{k!(n-k!)}$ for any integers $0 \le k \le n$




2. Show that $\binom{n-1}{k-1}+\binom{n-1}{k} = \binom{n}{k}$ for any $1 \le k \le n$

(I don't need help with this part, I have worked it out and it is true. This must be a hint for how to use induction, but I can't figure out exactly how to apply it, hmm...)

3. Now, using point (2) and induction, prove that for any integer $n \ge 1$ and any real number $x$,
   $$(1+x)^n = \sum_{k=0}^n x^k \binom{n}{k}$$

I'm guessing that the solution will require strong induction, i.e. I'll need to assume that for some integer a, the equivalence holds for all b in the range $1 \le b \le a$ and using this assumption show that the equivalence holds for $a+1$ as well. Perhaps multiply both sides by $(1+x)$? But that really messes up the binomial terms... Any help would be greatly appreciated! Thank you :)