y∈ℝ, y>−1 and y≠0. Show that for k∈ℕ, k>1: (1+y)^k > 1+ky

Question

Let $y∈ℝ$, $y>−1$ and $y≠0$. Show that for $k∈ℕ$, $k>1$ we have $(1+y)^k > 1+ky$.

I've been trying to solve this for the last two hours and still don't really have any idea on how to do it.

Answer 1

We will prove that $$(1+y)^k\geq 1+ky$$. For $k=1$ it is true. Now we assume that $$(1+y)^k\geq 1+ky$$ Now we will prove that $$(1+y)^{k+1}\geq 1+(k+1)y$$ Multiplying $$(1+y)^k\geq 1+ky$$ by $y+1>0$ and we get $$(1+y)^{k+1}\geq (1+ky)(1+y)=1+ky+y+ky^2$$ and now is is easy, that $$1+ky+y+ky\geq 1+(k+1)y$$ since $$ky^2\geq 0$$ It is Bernoulli's inequality