How to prove that $\log(x)<x$ when $x>1$?

You may just differentiate $$ f(x):=\log x-x, \quad x\geq1, $$ giving $$ f'(x)=\frac1x-1=\frac{1-x}x<0 \quad \text{for}\quad x>1 $$ since $$ f(1)=-1<0 $$ and $f$ is strictly decreasing, then $$ f(x)<0, \quad x>1, $$ that is $$ \log x -x <0, \quad x>1. $$

I thought it might be instructive to present a proof that relies on standard tools only. We begin with the limit definition of the exponential function

$$e^x=\lim_{n\to \infty}\left(1+\frac xn\right)^n$$

It is easy to show that the sequence $e_n(x)=\left(1+\frac xn\right)^n$ increases monotonically for $x>-1$. To show this we simply analyze the ratio

$$\begin{align} \frac{e_{n+1}(x)}{e_n(x)}&=\frac{\left(1+\frac x{n+1}\right)^{n+1}}{\left(1+\frac xn\right)^n}\\\\ &=\left(1+\frac{-x}{(n+x)(n+1)}\right)^{n+1}\left(1+\frac xn\right) \tag 1\\\\ &\ge \left(1+\frac{-x}{n+x}\right)\left(1+\frac xn\right)\tag 2\\\\ &=1 \end{align}$$

where in going from $(1)$ to $(2)$ we used Bernoulli's Inequality. Note that $(2)$ is valid whenever $n>-x$ or $x>-n$.

Since $e_n(x)$ monotonically increases and is bounded above by $e^x$, then

$$e^x\ge \left(1+\frac xn\right)^n \tag 3$$

for all $n\ge 1$. And therefore, for $x>-1$ we have

$$e^x\ge 1+x \tag 4$$

Since $e^x>0$ for all $x$, then $(4)$ is true for $x\le -1$ also. Therefore, $e^x\ge 1+x$ for all $x$.

ASIDE:

From $(4)$ we note that $e^{-x}\ge 1-x$. If $x<1$, then since $e^x\,e^{-x}=1$, $e^x\le \frac{1}{1-x}$. Thus, for $x<1$ we can write

$$1+x\le e^x\le \frac{1}{1-x}$$

Taking the logarithm of both sides of $(4)$ produces the coveted inequality

$$\log(1+x)\le x \tag 5$$

Interestingly, setting $x=-z/(z+1)$ into $(4)$ reveals

$$\log(1+z)\ge \frac{z}{z+1}$$

for $z>-1$. Putting it all together we have for $x>0$

$$\frac{x-1}{x}\le \log x\le x-1<x$$

If you defined the logarithm as $$\log(x)=\int_{1}^{x}{\frac{1}{t}dt},$$ $$\frac{1}{x} \le 1 \; \text{ for }x\ge 1.$$ Hence, $$ \log(x)=\int_{1}^{x}{\frac{1}{t}\,dt} \le \int_{1}^{x}\!{1}\,dt =x-1 \le x.$$ If $0< x\le 1\;$ then you simply get $$\log(x)=\int_{1}^{x}{\frac{1}{t}\,dt}=- \int_{x}^{1}{\frac{1}{t}\,dt}\le 0 < x.$$