How can I prove by induction that $$ \left( 1+\frac{5}{n} \right)^n < n $$ I have found that $n$ has to be at least 136.
When I try to use induction I only get $$ \left( 1+\frac{5}{n+1} \right)^{n+1} = \left( 1+\frac{5}{n+1} \right)^{n} \cdot \left( 1+\frac{5}{n+1} \right) = \left(\frac{1}{\frac{n+1}{n+6}}\right)^n\cdot\frac{n+6}{n+1} $$
Taking natural logs yields $n\log(1+5/n)$. This yields $$\dfrac{\log(1+5/n)}{1/n}.$$ Taking derivatives of the numerator and denominator yields
$$\dfrac{-5/(n^2+5n)}{-1/(n^2)}.$$ By division we get $$\dfrac{5}{1+5/n}.$$ By Algebraic limit properties $$\lim_{n\to\infty}\dfrac{5}{1+5/n}=5.$$ By L'Hopital it follows that $$\lim_{n\to\infty}\dfrac{\log(1+5/n)}{1/n}=5.$$ By continuity it follows that $$\lim_{n\to\infty}\text{exp}\bigg(\dfrac{\log(1+5/n)}{1/n}\bigg)=\lim_{n\to\infty}\bigg(1+\dfrac{5}{n}\bigg)^n=e^5.$$ The sequence has a finite limit, therefore is bounded, hence there exists an $N\in\mathbb{N}$ such that for all $n\geq N$ it follows that $$n>\bigg(1+\frac{5}{n}\bigg)^n.$$