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

Question

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

Please check whether it is $7n$ or $7^n$. Your question and heading didn't match — Swapnil Tripathi, Oct 27 '14 at 20:45
yes but i dont understand, base case is when n=21, is it true for base case? — Amirpasha, Oct 27 '14 at 20:57
@SwapnilTripathi Do you know if we can verify the base case easily (w/o calculator)? — Akiva Weinberger, Oct 27 '14 at 21:07

Swapnil Tripathi · Accepted Answer · 2014-10-27T21:48:08.210

3

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

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

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

$21!>7^{21}$

edited Oct 27 '14 at 21:48

answered Oct 27 '14 at 21:01

Swapnil Tripathi

I can rewrite 7^k+1 as 7^k * 7 – Amirpasha Oct 27 '14 at 21:03
Yes. And $7^k\cdot 7<k!\cdot 7<k!\cdot(k+1)$. That is because $k\ge 21>7$ and so is $k+1$. Were you able to show this for the base case or do you need help? – Swapnil Tripathi Oct 27 '14 at 21:06
thank you very much for your kind help))) – Amirpasha Oct 27 '14 at 21:54
You're welcome. I got stuck somewhere while proving and it took me a while to know where I was going wrong – Swapnil Tripathi Oct 27 '14 at 21:56

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