Prove by induction that $n!>2^n$

Suppose that when $n=k$ $(k≥4)$, we have that $k!>2^k$.

Now, we have to prove that $(k+1)!>2^{k+1}$ when $n=(k+1) (k≥4)$.

$(k+1)! = (k+1)k! > (k+1)2^k$ (since $k!>2^k$)

That implies $(k+1)!>2^k \cdot 2$ (since $(k+1)>2$ because of $k$ is greater than or equal to $4$)

Therefore, $(k+1)!>2^{k+1}$

Finally, we may conclude that $n!>2^n$ for all integers $n≥4$


Here's a suggestion. We have that $(n+1)! = (n+1)n!$ and $2^{n+1} = 2\cdot 2^n$. Then, if we know that $n! > 2^n$, and we multiply $n!$ by $n+1$ and $2^n$ by $2$, can you work out what will happen to the inequality?


Guide: If $a>b>0$ and $c>d>0$, then we know that $ac>bd$. Now if we know $n!>2^n$ for some positive integer $n$, and we also know that $n+1>2$....

Tags:

Inequality

Factorial

Induction

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