real analysis - prove $log(1+x)0$

Prove $\log(1+x) \lt x$ for $x\gt0$

my attempt:

I show $e^{x}\gt 1+x$ for $x\gt0$
since

$e^{x}=1+x+\frac{n(n-1)}{2}\frac{x^2}{n^2}+...$

so if $x\gt0$ then all terms are positive

so $e^{x}\gt 1+x$ for $x\gt0$

now given $e^{x}\gt 1+x$ for $x\gt0$, can I take $\log$ on both sides and show

$\log(1+x)\lt x$ for $x\gt0$

or do I have to prove firstly that $\exp(x)=e^x$ for $x\gt 0$ then I can take $\log$..