Prove by the method of Mathematical induction that $(1-0.3)^n \geq 1-0.3n$ for all $n$ in set of positive integers

Question

Here is what I have so far

Basis

For $n = 0 (1-0.3)^0 \geq 1-0.3(0)$ checks

For $n = 1 (1-0.3)^1 \geq 1-0.3(k$) checks

I.H. $(1-0.3)^k \geq 1-0.3(k)$ for all k in the set of positive integers (1)

We want to prove that $(1-0.3)^{k+1} \geq 1-0.3(k+1)$ (2)

To relate (1) to (2) we have to add $(1-0.3)^{k+1}$ to both sides of (1).

$(1-0.3)^k+(1-0.3)^{k+1} \geq 1-0.3(k) + (1-0.3)^{k+1}$.

Here is where I am stuck. I need help as to where to go from here.

Answer 1

I don't see why you have to add $(1-0.3)^{k+1}$.

\begin{align} (1-0.3)^{k+1} &= (1-0.3)(1-0.3)^k \\ &\geq (1-0.3) (1-0.3k) \\ &= 1-0.3-0.3k+0.3^2k \\ &=1-0.3(k+1)+0.3^2k \\ &\geq 1-0.3(k+1) \end{align}