Proof using induction n!

Question

I wanna know how to proof using induction; I saw this example in a discrete book; however, i could not solve it; the question is below:

Prove, using induction, that $3^{n} < n!$ for all $n ≥ 7$

Answer 1

Initial case: The case clearly holds for $n = 7$. (Easily checked with a calculator)

Inductive hypothesis: $3^{n-1} < (n-1)!$

Inductive step: $$ 3^n = 3 \cdot 3^{n-1} < 3 (n-1)! < n (n-1)! = n! $$

I think the purpose of this problem is to demonstrate how to use induction, instead of the importance of the fact $3^n < n!$.

Answer 2

I'm not sure what "induction" is ( i'm not an English native speaker ) but that's how I would do it :

First take the case of $n = 7$

$3^7= 2187$ and $7! = 5040 $

so it's true for $n=7$

Then suppose for some $n \geq 7$ that we have $3^n < n! $

now we multiply both sides of by 3 and get

$3^{n+1} < 3(n!)$

and we know $n \leq 7$ so

$3(n!) < (n+1)! $

So if we combine our two above steps we get:

$3^{n+1}< (n+1)!$

There you go! It's true for $n=7$ and if true for any $n \geq 7$ then it's true for $n+1$ so you can deduce it's true for any $n \geq 7$.

Answer 3

First, show that this is true for $n=7$:

$3^7<7!$

Second, assume that this is true for $n$:

$3^n<n!$

Third, prove that this is true for $n+1$:

$3^{n+1}=$

$3\cdot\color\red{3^n}<$

$3\cdot\color\red{n!}<$

$7\cdot{n!}<$

$(n+1)\cdot{n!}=$

$(n+1)!$

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Please note that the assumption is used only in the part marked red.