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

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

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

Yes, one can use $$\tag1e^x\ge 1+x,$$ which holds for all $x\in\mathbb R$ (and can be dubbed the most useful inequality involving the exponential function). This again can be shown in several ways. If you defined $e^x$ as limit $\lim_{n\to\infty}\left(1+\frac xn\right)^n$, then $(1)$ follows from Bernoullis inequality: $(1+t)^n>1+nt$ if $t>-1$ and $n>0$.

To show that $\ln(x)\le x-1$ for all $x>0$, just substitute $\ln x$ for $x$ in $(1)$.

$y=x-1$ is the equation of the tangent to the ln curve at $(1,0)$ and the function is concave, hence its graph is under the tangent.