Prove that $1+rx = (1+x)^r$ holds only if $x=0$. I believe my work is correct here, just looking for closure. Is it true  
that since r>1, the last line could be (1+x)^1? 


Suppose $1+rx\le(1+x)^r$  for any real number $r>1$ and $x>-1$. Prove that the equality holds only if $x=0$.  

My work: For $x=0$, letting $f(x) = (1+x)^r$: [mean value theorem]
$f'(c) = \frac{f(x)-f(0)}{x-0}$  
$r(1+c)^{r-1} = \frac{(1+x)^r-1^r}{x-0}$ 
$\frac{(1+x)^r-1^r}{(1+c)^{r-1}}=xr$
$(1+x)-1=xr$
$1+x = 1+xr$

Let $f(x)=(1+x)^r-rx-1$ then $f'(x)=r(1+x)^{r-1}-r>0$ for $x\neq 0$ since $r-1>0$, so $f$ is monotonically increasing and $f(0)=0$ hence $f(x)\neq 0$ for all $x\neq 0$. We can conclude.
