Using induction, how to find all positive number $n$ such that $5^{n} > n!$?

Question

Using induction, how to find all positive number $n$ such that $$ 5^{n} > n!$$

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Attempt:

I found this problem in the Induction topic in an education material.

I know starting that $n=12$ then $5^{n} < n!$. Is it because that we have to prove that for $n \ge 12$ then $5^{n} < n!$ by induction...? Hence the answer is $n=1,2,3,4,5,6,7,8,9,10,11$.

Or is there another way using induction.

Answer 1

As you have observed, $5^n > n!$ only if $n\in\{1,2,3,4,5,6,7,8,9,10,11\}$. Induction will show that $5^n < n!$ with $n=12$ as your base step.

Base Case:

If $n=12$, then $5^{12}<(12)!$.

Induction Hypothesis:

Assume that $5^n < n!$ for all $n\in\mathbb N$ where $n\ge 12$.

Induction Step:

We need to show that $5^{n+1} < (n+1)!$ for all $n\in\mathbb N$ where $n\ge 12$. Starting from the LHS

\begin{align}5^{n+1}&=5^n5 \\&<n!\cdot 5 \\&< n!\cdot n \\&< n!\cdot (n+1) \\&=(n+1)!\end{align}

Therefore, this allows us to conclude that

1. $5^n > n!$ for $n\in\{1,2,3,4,5,6,7,8,9,10,11\}$.
2. $5^n < n!$ for $n\in\mathbb N$ and $n\ge 12$.