Question

n!>2^n for all integers n≥4

Answer 1

Take logarithm on both numbers.

$\text{ \cdots\longrightarrow\log{n!}\text{ or }\log{2^{n}} }$

By properties of logarithms,

$\cdots\longrightarrow\log{1}+\log{2}+\cdots+\log{n}\text{ or }n\log{2}$

Let's divide both sides by $\log{2}$.

$\text{ \cdots\longrightarrow\dfrac{\log{1}}{\log{2}}+\dfrac{\log{2}}{\log{2}}+\cdots+\dfrac{\log{n}}{\log2}\text{ or }n }$

We know that $\largen!$ exceeds $\large2^{n}$ at $\largen=4$. Assume we know the base change of logarithms.

$\cdots\longrightarrow\log_{2}{1}+\log_{2}{2}+\cdots+\log_{2}{n}\text{ or }n$

Putting $n=4$, -

$\cdots\longrightarrow\log_{2}{1}+\log_{2}{2}+\cdots+\log_{2}{4}>4$

For $\largek\geq2$,

$\cdots\longrightarrow\log_{2}{k}>k.$

Putting $n=k+1$, -

$\cdots\longrightarrow\log_{2}{1}+\log_{2}{2}+\cdots+\log_{2}{k}>k+1.$

By mathematical induction, the inequality $\largen!>2^{n}$ is true for $\largen\geq4$ $\blacksquare$