Not understanding the logic behind $2<k+1$ in Induction Proof Problem

Question

$2^k+1 = 2 · 2^k$

$< 2 · k!$

$< (k + 1)k!$ (because $2 < k + 1$)

$= (k + 1)!$ (by definition of factorial function.)

I am not getting the third line. where did this $2 < k + 1$ appear.

This is from the book Discrete Mathematics and Its Application by Kenneth H Rosen chapter 5.1 Mathematical Induction (7th ed) Example 6.

Full Problem and it's Solution given below:

Use mathematical induction to prove that $2^n < n!$ for every integer $n$ with $n ≥ 4$. (Note that this inequality is false for $n = 1, 2, and$ $3$.)

Solution: Let $P(n)$ be the proposition that $2^n < n!$.
BASIS STEP: To prove the inequality for $n ≥ 4$ requires that the basis step be $P (4)$. Note that $P (4)$ is true, because $2^4 = 16 < 24 = 4!$

INDUCTIVE STEP: For the inductive step, we assume that $P (k)$ is true for an arbitrary integer $k$ with $k ≥ 4$. That is, we assume that $2^k < k!$ for the positive integer $k$ with $k ≥ 4$. We must show that under this hypothesis, $P (k + 1)$ is also true. That is, we must show that if $2^k < k!$ for an arbitrary positive integer $ k$ where $k ≥ 4$, then $2^k+1 < (k + 1)!$. We have
$2^k+1 = 2 · 2^k$ (by definition of exponent)
$< 2 · k!$ (by the inductive hypothesis)
$< (k + 1)k! $ (because $2 < k + 1$)
$= (k + 1)!$ (by definition of factorial function.)
This shows that $P (k + 1)$ is true when $P (k)$ is true. This completes the inductive step of the proof. We have completed the basis step and the inductive step. Hence, by mathematical induction $P (n)$ is true for all integers $n$ with $n ≥ 4$. That is, we have proved that $2^n < n!$ is true for all integers $n$ with $n ≥ 4$.

Answer 1

You want to show by induction that $2^n < n!$ for $n\geq 4$. Induction consists of two parts: The start and the induction step.

We start with the inductions start where we just verify that the statement is true for the lowest $n$ possible, here we have $n=4: 2^4 = 16 < 4! = 24$. That is indeed correct.

Now we prove the induction step. To not confuse the notation another variable $k$ is used. We assume that the above statement is correct for some arbitrary $k\geq 4$. We want to show that - given this condition - the statement is true for $k+1$ as well. So our goal is to show that $$ 2^{k+1}< (k+1)! $$

After some simplification we get to the point you mentioned: $2\cdot k! < (k+1)\cdot k!$

The inequality $2<k+1$ simply comes from the property of the arbitrary chosen $k$ which is already greater or equal to $4$. So $k+1\geq 5 > 2$ and the proof is completed.