Understanding a step in this proof by induction Here's an example I discovered in a book.

Prove inequality when $a\ge-1$:$$(1 + a)^n \ge 1 + na.$$
Let's use mathematical induction. Then $n = 1$ left and right parts are equals.
Let's suppose statemant is right when $n = k.$ Multiply on $(1 + a)$: $$(1 + a)^{k + 1}\ge (1 + ka)(1 + a)\\(1 + a)^{k + 1} \ge 1 + (k + 1)a + ka^2$$
Removing $ka^2$ as non-negative in right part of equation get: $$(1 + a)^{k + 1} \ge 1 + (k + 1)a$$ Inequality is proved.

I don't understand: why they can remove $ka^2$ in right part of equation?
 A: The author is leaving out (arguably tedious) steps. This may be because the author thinks they are obvious, and so not worth mentioning. This may be because the author thinks they are not too far from obvious, and so leaves the additional steps as an understanding-enhancement exercise for the reader.
Note that $k$ is positive and $a^2\ge0.$ (Do you see why both of these claims are true?) Hence, since $$0\le a^2,$$ then by positivity of $k$ we have $$k\cdot 0\le k\cdot a^2,$$ which can be rewritten as $$0\le ka^2.\tag{1}$$ (With me so far?) Now, adding $1+(k+1)a$ to both sides of $(1)$, we have $$1+(k+1)a\le 1+(k+1)a+ka^2,$$ which can be rewritten as $$1+(k+1)a\le (1+ka)(1+a).\tag{2}$$ Now, since $-1\le a,$ then $0\le 1+a.$ From this, and the assumption that $$1+ka\le(1+a)^k,$$ we can conclude that $$(1+ka)(1+a)\le(1+a)^k(1+a),$$ which can be rewritten as $$(1+ka)(1+a)\le(1+a)^{k+1}\tag{3}.$$ (Still with me?) Finally, from $(2)$ and $(3),$ we conclude that $$1+(k+1)a\le(1+a)^{k+1},\tag{$\heartsuit$}$$ as desired.
If you have trouble justifying any of these steps (or even if you just want to bounce proof attempts off of somebody), feel free to leave a comment, and I will reply as soon as I am able.
