Let $I$ be the interval $(-1, \infty)$ and $f :I \rightarrow \mathbb R$ the function defined as:$$ f(x) = (1 + x)^\alpha - 1 - \alpha x$$ with $\alpha \in \mathbb R \setminus \lbrace 0, 1 \rbrace$ fixed. Then $f$ is differentiable and its derivative is$$ f'(x) = \alpha (1 + x)^{\alpha - 1} - \alpha, \mbox{ for all } x \in I,$$ from which it follows that $f'(x) = 0 \Leftrightarrow x = 0$ .

1. If $0 < \alpha < 1$ then $f'(x) < 0$ for all $x \in (0, \infty)$ and $f'(x) > 0$ for all $x \in (-1, 0)$ which means that $0$ is a global maximum point for $f$ . Therefore $f(x) < f(0)$ for all $x \in I \setminus \lbrace 0 \rbrace$ which means that $(1 + x)^\alpha < 1 + \alpha x$ for all $x \in (-1, 0)$ .
2. If $\alpha \notin [0, 1]$ then $f'(x) > 0$ for all $x \in (0, \infty)$ and $f'(x) < 0$ for all $x \in (-1, 0)$ meaning that $0$ is a global minimum point for $f$ . This implies that $f(x) > f(0)$ for all $x \in I \setminus \lbrace 0 \rbrace$ which means that $(1 + x)^\alpha > 1 + \alpha x$ for all $x \in (-1, 0)$ .

Checking that the equality is satisfied for $x = 0$ or for ends the proof.