How to prove $(1+q)^n \geq1+qn$ for all $\mathbb{N}$ with $q>0$
Solution 1:
You can actually do the whole same process by admitting that $q>-1$. The calculations could be organized a little more neatly:
$$\begin{align*} (1+q)^{k+1}&=(1+q)(1+q)^k\\ &\ge(1+q)(1+kq)\\ &=1+(k+1)q+kq^2\\ &\ge 1+(k+1)q\;, \end{align*}$$
since $kq^2\ge 0$. This completes the induction step. This result is known as the Bernoulli's inequality.