If $a,x\in \mathbb{R}$, I want to prove that $(1+x)^a$ is bounded by $b\cdot x^a$ if $x>1$, where $b$ is a constant independent of $x$. I believe that Newton bynomial is somehow implicit in the proof I have to do. This is because, if $a\in\mathbb{Z}$, $$ (1+x)^a=\sum_{i=0}^{a}{a\choose i}x^{i}. $$ So, $$ (1+x)^a=\sum_{i=0}^{a}{a\choose i}x^{i}\leq b\cdot x^{a} $$ where $b=\sum_{i=0}^{a}{a\choose i}$. Nevertheless, my $a$ is not neccesarily entire. So, my attempt has been to use the floor function of $a$ in some way... For example, by saying that $(1+x)^a\leq (1+x)^{\lfloor a\rfloor +1}$ and applying Newton to the upper bound. But, I am not arriving to my desire fact.
Any advice will be highly appreciated, thanks!