By differentiating the nth binomial expansion, it is possible to deduce the (n-1)th expansion is true. Is it possible to then show that the nth expansion implies the k*nth expansion for some positive integer k?
Second question: Can the binomial theorem be proven by integrating the nth expansion?
You can almost get the answer, albeit with some cheating.
First, $(x+y)^n=x^n(1+(y/x))^n$, so we can reduce this to showing:
$$(1+x)^n = \sum_{k=0}^n\binom{n}{k}x^k$$ $$(1+x)^{n+1}=(n+1)\int_{-1}^x(1+t)^{n}dt=\sum_{k=0}^n\binom{n}{k}\frac{n+1}{k+1}(x^{k+1}-(-1)^{k+1}).$$
$$\binom{n}{k}\frac{n+1}{k+1}=\frac{(n+1)!}{(k+1)!(n+1-(k+1))!}=\binom{n+1}{k+1}.$$
$$\sum_{k=0}^n\binom{n}{k}\frac{n+1}{k+1}(x^{k+1}-(-1)^{k+1})=\sum_{k=1}^{n+1}\binom{n+1}{k}x^{k}+1=\sum_{k=0}^{n+1}\binom{n+1}{k}x^k,$$
where we reindexed the sum and used the fact that $(1-1)^{n+1}=0$. This last bit cheats in a way because we need to know that $(1-1)^{n+1}=\sum_{k=0}^n\binom{n+1}{k}(-1)^k$.