$\abs{E_p(z)- 1} \le \abs z^{p + 1}$

hbghlyj

$p\inN^+$，$z\inC$，$E_p(z) = (1 - z) \exp\left(z + \dfrac {z^2} 2 + \cdots + \dfrac {z^p} p\right)$，
如何证明这里的Another bound

Let $\abs z \le 1$. Then:
$$\abs{E_p(z) - 1} \le \abs z^{p + 1}$$

This theorem requires a proof.