Prove that e^x> 1+x for all x ≠0

Prove that $e^{x}\gt 1+x$ for all $x\neq0$.

Step 1: Reformulate

\[ f(x) = e^{x}-1-x, \qquad x\in\mathbb{R}; \] the claim is that $f(x)\gt 0$ for every $x\neq0$.

Step 2: The only critical point

\[ f'(x) = e^{x}-1 = 0 \;\Longleftrightarrow\; x = 0. \]

Step 3: It is a global minimum

\[ f''(x) = e^{x}\gt 0 \quad\text{for every } x, \] so $f$ is strictly convex on the whole line and its single critical point is the global minimum. Equivalently, $f'\lt 0$ for $x\lt 0$ and $f'\gt 0$ for $x\gt 0$.

Step 4: The minimum value

\[ f(0) = e^{0}-1-0 = 0. \]

Strict convexity makes the minimum strict, so $f(x)\gt 0$ for every $x\neq0$.

Step 5: Verification

\[x = 1:\ e = 2.7182818 \gt 2; \qquad x = -1:\ e^{-1} = 0.3678794 \gt 0 \;\;\checkmark\]

\[x = 0.1:\ 1.1051709 \gt 1.1; \qquad x = -0.1:\ 0.9048374 \gt 0.9 \;\;\checkmark\]

\[\text{Series: } e^{x}-1-x = \frac{x^{2}}{2}+\frac{x^{3}}{6}+\cdots,\ \text{positive for small } x \text{ of either sign} \;\;\checkmark\]