inequality - Show that $ln (x) leq x-1$

Show that $\ln (x) \leq x-1$

I'm not really sure how to show this, it's obvious if we draw a graph of it but that won't suffice here. Could we somehow use the fact that $e^x$ is the inverse? I mean, if $e^{x-1} \geq x$ then would the statement be proved?

Answer

Define for $\;x>0\;$

$$f(x)=\ln x-x+1\implies f'(x)=\frac1x-1=0\iff x=1$$

and since $\;f''(x)=-\dfrac1{x^2}<0\quad \forall x>0\;$ , we get a maximal point.

But also

$$\lim_{x\to 0+}f(x)=-\infty=\lim_{x\to\infty}f(x)$$

Thus, the above is a global maximal point and

$$\forall\,x>0\;,\;\;\;f(x)\le f(1)=0$$