Let us take as our assumption that $x \in I = \left(-1, \infty\right)$ and that $r \in J = \left(0, \infty\right)$ . Observe that if $x = 0$ the inequality holds quite obviously. Let us now consider the case where $x \neq 0$ .
Consider now the function $f: I{x}J\rightarrow\mathbb{R}$ given by $$f(x,r) = (1 + x)^r - 1 - rx$$ Observe that for all $r$ in $J$ fixed, $f$ is, indeed, differentiable on $I$ . In particular, $$\frac{\partial}{{\partial}x}f(x,r) = r(1+x)^{r-1}-r$$ Consider two points $a \neq 0$ in $I$ and $0$ in $I$ . Then clearly by the mean value theorem, for any arbitrary, fixed $\alpha$ in $J$ , there exists a $c$ in $I$ such that, $$f'_x(c, \alpha) = \frac{f(a, \alpha) - f(0, \alpha)}{a}$$ \begin{equation} \Leftrightarrow f'_xf(c, \alpha)=\frac{(1+a)^\alpha-1-{\alpha}a}{a}\label{eq:last} \end{equation}Since $\alpha$ is in $J$ , it is clear that if $a < 0$ , then $$f'_x(a,\alpha) < 0$$ and, accordingly, if $a > 0$ then $$f'_x(a,\alpha) > 0$$ Thus, in either case, from we deduce that $$\frac{(1+a)^\alpha-1-{\alpha}a}{a} < 0$$ if $a < 0$ and $$\frac{(1+a)^\alpha-1-{\alpha}a}{a} > 0$$ if $a > 0$ . From this we conclude that, in either case,$(1+a)^\alpha-1-{\alpha}a > 0$ . That is, $$(1 + a)^\alpha > 1+{\alpha}a$$ for all choices of $a$ in $I - \left\{0\right\}$ and all choices of $\alpha$ in $J$ . If $a = 0$ in $I$ , we have$$(1 + a)^\alpha = 1+{\alpha}a$$ for all choices of $\alpha$ in $J$ . Generally, for all $x$ in $I$ and all $r$ in $J$ we have: $$(1+x)^r \geq 1 + rx$$ This completes the proof.

Notice that if $r$ is in $\left(-1, 0\right)$ then the inequality would be reversed. That is: $$(1+x)^r \leq 1 + rx$$ . This can be proved using exactly the same method, by fixing $\alpha$ in the proof above in $\left(-1, 0\right)$ .