Prove $\ln(n) \lt n$ using only $\log(x^y) = y\log(x) $

Question

Prove $$\ln(n) \lt n \;\;(n\in \mathbb N) $$ Using only the rule $$\log(x^y) = y\log(x) $$ I tried using any of the known Inequalities am-gm, power mean, titu's lemma, holder's,... but I seem I can't go anywhere near required. Any hint?

$\mathbf {Edit} $
It was simpler than I thought, it's done with Bernoulli's Inequality. $$e^n\gt 2^n = (1+1)^n \ge 1+n \gt n$$ $$\ln(e^n) \gt \ln(n)$$ $$\ln(n) \lt n$$

Answer 1

Your inequality could be generalized as $$\ln x \leq x-1, x>0$$ To prove it, you perhaps have to use calculus, not just elementary mathematics.Notice that, for $f(x)=\ln x-x+1,(x>0)$, $$f'(x)=\frac{1}{x}-1.$$ Thus, $f'(x)>0,$ when $0<x<1$; $f'(x)=0$, when $x=1$; $f'(x)<0$, when $f'(x)>1.$ For the reason, $f(x)$ reaches its maximum value $f(1)=0$ at $x=0$, namely, $f(x) \leq 0$.

But if you have known Bernoulli's inequality, there indeed exists an elementary proof.

Since

$$(1+x)^n \geq 1+nx,(x \geq -1, n\in \mathbb{N_+}).$$

Let $x=e-1$. Then $$e^n \geq 1+n(e-1)=1+ne-n>n.$$

It follows that $$n>\ln n.$$

Answer 2

You cannot prove $\log(n) \lt n$ just using $\log(x^y) = y\log(x)$.

Let $k={\sqrt[3]{2}} \approx 1.259921$.

Then $\log_k(x^y) = y\log_k(x)$ as required but $\log_k(2) = 3 \not\lt 2$