## Algebra and Trigonometry 10th Edition

The inequality was proved for $n=1$ The inequality is correct if $n$ is changed by $n+1$
Let's prove the inequality for $n=4$: $4!\gt2^4$ $4\times3\times2\times1\gt16$ $24\gt16$ It is correct! Now, suppose that the inequality is correct, that is: $n!\gt2^n$ for $n\geq4$ Now, let's prove the inequality for $n+1$: $(n+1)!=(n+1)n(n-1)(n-2)...3\times2\times1=(n+1)n!$ Since $n\geq4$ we have that $n+1\geq5\gt2$: $(n+1)!=(n+1)n!\gt2(n!)\gt2(2^n)=2^{n+1}$ $(n+1)!\gt2^{n+1}$ That is exactly the given inequality if $n$ is changed by $n+1$