real analysis - Using the Law of the Mean (MVT) to prove the inequality $log(1+x)

If $x \gt0$, then $\log(1+x) \lt x$.

My attempt at the proof thus far...

Let $f(x) = x-\log(1+x)$, then $f'(x) = 1-\frac{1}{1+x}$ for some $\mu \in (0,x)$ (since $x>0$)

MVT give us $f(x) = f(a) + f'(\mu)(x-a)$

So plug in our values we get:

$$x-\log(1+x) = 0+(1-\frac{1}{1+\mu})(x-0)$$

which we can reduce to

$$\log(1+x)=\frac{x}{1+\mu}$$

Now For $x\leq1$, then $0\lt\mu\lt1$

S0 $0\lt \frac{1}{1+\mu}\lt 1$, thus $\log(1+x)\lt x$

If $x>1$, then....

So I can see clearly that if $x>1$ is plugged into $\frac{x}{1+\mu}$

then $\log(1+x)

I would appreciate tips hints or proof completions.