Mathematical induction with an inequality involving factorials

Question

How do I prove this inequality: $n!>2^n$, where $n\ge 4$. I've searched online but the solutions are vague to me. Thank you! :)

Answer 1

A proof by induction has three parts: a basis, induction hypothesis, and an inductive step. We show that the basis is true, and then assume that the induction hypothesis is true. We then use our assumption to imply this inequality is true for all other values.

Basis: Let $n=5$. Then $n!=5!=120$. $2^n=2^5=32$. $120>32$.

Induction hypothesis: Suppose $n=k>4$. Assume that $k!>n^k$ holds true.

Inductive: Now let $n=k+1$. $(n+1)!=(n+1)n!$. $2^{k+1}$= $2^k2$. We know $n!>2^k$, so now we must simply compare $n+1$ and $2$. $n$ is strictly greater than $4$, so $n+1$ is certainly greater than $2$. Thus $(n+1)!>2^{k+1}$.

Thus we have shown by induction that $n!>2^k$

Answer 2

We will prove by induction

Base case $n=5$ , $5!=120>32=2^5$

Case $n=k$ assuming true for $n=k-1$

$(k-1)! >2^{(k-1)}$ multiplying by $k$ yields

$k(k-1)! >k*2^{(k-1)}> 2*2^{(k-1)}=2^k$ since $k>2$

$k! >2^k$

Answer 3

Assume $$2^n<n!$$ multiply by $2$ to get $$2^{n+1}<2n!$$ and we see that $$2n!\leq(n+1)n!=(n+1)!$$