Proving that $3^n<n!$ when $n\geq 7$

Question

It's been 10 years since my last math class so I'm very rusty. How would I go about proving $$3^n < n!$$ where $n \geq 7$?

I understand that factorials grow faster than set values with a variable exponent. Just not sure how to start proving it mathematically.

Answer 1

We can prove it by induction. For $n\geq 7$, let $S(n)$ denote the statement $$ S(n) : 3^n < n!. $$ Base case ($n=7$): $S(7)$ says that $3^7 = 2187<5040=7!$, and this is true.

Induction step: Fix some $k\geq 7$, and assume that $S(k)$ is true where $$ S(k) : 3^k < k! $$ To be shown is that $S(k+1)$ is true where $$ S(k+1) : 3^{k+1} < (k+1)! $$ Beginning with the left-hand side of $S(k+1)$, \begin{align} 3^{k+1} &= 3^k\cdot3\tag{by definition}\\[0.5em] &< k!\cdot 3\tag{by $S(k)$, the ind. hyp.}\\[0.5em] &< k!\cdot (k+1)\tag{since $k\geq 7$}\\[0.5em] &= (k+1)!, \end{align} we end up at the right-hand side of $S(k+1)$, completing the inductive step.

By mathematical induction, the statement $S(n)$ is true for all $n\geq 7$. $\blacksquare$

Answer 2

Since $2187=3^{7}<7!=5040$ then for $n>7$ $$ 3^{n}=3^{7}\cdot3^{n-7}<7!\cdot8\cdot9\cdot\ldots\cdot n=n! $$

since $8,9,\dots n>3$