I'm reading through example #3 on Purplemath. I don't understand the reasoning behind the line I marked $\dagger$ and why the underlined term, $\underline{2^k}$ gets added!? I'm rewriting the example:
(*) Prove that for $n \geq 5, 4n < 2^n$.
This one doesn't start at $n = 1$, and involves an inequality instead of an equation. [...]
BASE CASE:
Let $n = 5$.
Then $4n = 4×5 = 20$, and $2^n = 2^5 = 32$.
Since $20 < 32$, then (*) holds at $n = 5$.
INDUCTION STEP:
Assume, for $n = k$, that (*) holds; that is, assume that $4k < 2^k$.
Let $n = k + 1$
The left-hand side of (*) gives us $4(k + 1) = 4k + 4$, and, by assumption,
$[4k] + 4 < [2^k] + 4$
PROBLEM AREA:
$\dagger$Since $k \geq 5$, then $4 < 32 \leq 2^k$. Then we get
$2^k + 4 < 2^k + \underline{2^k} = 2×2^k = 2^1×2^k = 2^{k+1}$
Then $4(k+1) < 2^{k+1}$, and (*) holds for $n = k + 1$.
Then (*) holds for all $n \geq 5$.
Because if $k\geq 5$ is an integer, $2^k\geq 32$ (exponential functions are strictly increasing when base $\gt 1$)
Now, you know from elementary math that $a\gt b\implies c+a\gt c+b$ for real $c$
So, you have, for $k\geq 5$ in the inductive step, $2^k\geq 32\gt 4$ and hence $2^k+2^k\gt 2^k+4$ or equivalently $2^k+4\lt 2^k+2^k$.
The motivation to add this stuff in particular is to complete the inductive step as already mentioned in
gt6989b's answer.