Given
$$\exp: \mathbb{R} \ni x \mapsto \sum_{k=0}^{\infty } \frac{1}{k!} x^{k} \in \mathbb{R}$$
also $e = \exp(1)$. For all $x \in \mathbb{R}$ with $\left | x \right | \leq 1$:
$$\left | \exp(x) - 1 \right | \leq \left | x \right | \cdot (e-1)$$
What would be the best way to proof this? By induction would work / be good? I would have to show then 2 things: That it's 1. smaller 2. equal? But not sure if it will work because of the exp(x) in the equation. Probably will work but end up too complicated?
Another idea but not sure, show that $\left | \exp(x) - 1 \right | \leq \sum_{k=1}^{\infty } \frac{1}{k!}\left | x^{k} \right |$ and use this to show that $\left | \exp(x) - 1 \right | \leq \left | x \right |* (e-1)$?
What would be the easiest way?
$$ \lvert \exp{x}-1 \rvert = \left\lvert \sum_{k=1}^{\infty} \frac{x^k}{k!} \right\rvert \leqslant \lvert x \rvert \sum_{k=1}^{\infty} \frac{\lvert x \rvert^{k-1}}{k!} \leqslant \lvert x \rvert \sum_{k=1}^{\infty} \frac{1}{k!} = \lvert x \rvert (e-1), $$ where the first inequality is the triangle inequality, and the second uses that $\lvert x \rvert \leqslant 1$.