Can you use mathematical induction to prove that $5^{n} < n!$ $5^n < n!$ for all $n in ZZ^+, n>=12$ ?

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Answer 1 · 2017-02-03T07:32:02

See explanation...

Proposition

Let $P(n)$ be the proposition:

$5^color(blue)(n) < color(blue)(n)!$

$color(white)()$
Base case

$P(color(blue)(12))$ is true since:

$5^color(blue)(12) = 244140625 < 479001600 = color(blue)(12)!$

$color(white)()$
Induction step

Suppose $P(color(blue)(n))$ is true:

$5^(color(blue)(n)) < color(blue)(n)!$

Then $P(color(blue)(n+1))$ is true:

$5^(color(blue)(n+1)) = 5*5^n < 5*n! < (n+1)*n! = (color(blue)(n+1))!$

$color(white)()$
Conclusion

Having shown:

${ (P(color(blue)(12))), (P(color(blue)(n)) => P(color(blue)(n+1)) " for " n >= 12) :}$

We can deduce:

$P(n)$ for all $n >= 12$

Can you use mathematical induction to prove that 5^n < n!5n<n!5^n < n! for all n in ZZ^+, n>=12n in ZZ^+, n>=12?