$$(\text b)\ \left\vert\exp(z)-\sum_{k=0}^n\frac{z^k}{k!}\right\vert=\left\vert\sum_{k=n+1}^\infty\frac{z^k}{k!}\right\vert\le\sum_{k=n+1}^\infty\frac{\vert z\vert^k}{k!}=\frac{\vert z\vert^{n+1}}{(n+1)!}\cdot\sum_{k=n+1}^\infty\frac{\vert z\vert^{k-(n+1)}}{\frac{k!}{(n+1)!}}\\=\frac{\vert z\vert^{n+1}}{(n+1)!}\cdot\sum_{k=0}^\infty\frac{\vert z\vert^k}{\frac{(k+n+1)!}{(n+1)!}}=\frac{\vert z\vert^{n+1}}{(n+1)!}\cdot\sum_{k=0}^\infty\frac{\vert z\vert^k}{(n+2)\cdots(n+k+1)}\\\le\frac{\vert z\vert^{n+1}}{(n+1)!}\cdot\sum_{k=0}^\infty\frac{\vert z\vert^k}{(n+2)^k}\le\frac{\vert z\vert^{n+1}}{(n+1)!}\cdot\sum_{k=0}^\infty\left(\frac12\right)^{\!k}\color{#999}{\text{, wenn }\vert z\vert<1+\tfrac n2}\\=\frac{2\cdot\vert z\vert^{n+1}}{(n+1)!}.$$
