Prove by induction that $7^n < n!$ for all integer $n \ge 21$

Question

Prove by induction that $7^n < n!\,$ for all integers $n\ge 21$

Answer 1

Step 1: Is $7^{21}<21!$ ? If yes, move to step 2.

Step 2: Suppose this holds for some $k(\ge 21)\in \mathbb{N}$

Now you need to show that if the condition holds for $k$, it also holds for $k+1$

Step 3: Easy-peasy. $7^{k}<k!$ holds and you need to show $7^{k+1}<(k+1)!$. What should you do?

Edit For Step 1

$\prod_{n=14}^{21}n>(14)^7\cdot21=2^7\cdot 3\cdot 7^8$

Also, $\prod_{n=7}^{13}n>7^6\cdot 13$

Combining both equations

$\prod_{n=7}^{21}n>2^7\cdot 3\cdot 7^{14}\cdot 13$

or

$21!=\prod_{n=1}^{21}n>1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 2^7\cdot 3\cdot 7^{14}\cdot 13$

Rearranging the terms

$21!=\prod_{n=1}^{21}n>(1\cdot2^3)\cdot (2\cdot 2^2) \cdot (3\cdot 3)\cdot (4\cdot 2)\cdot (5\cdot 2)\cdot 6\cdot 7^{14}\cdot 13$

$21!>6\cdot 7^{19}\cdot 13$

$21!>7^{21}$