Proof

We can estimate the remainder term of the exponential series as follows

By the definition of the complex exponential series, we have $$|r_{N + 1}(z)|= \sum_{n=N+1}^\infty\frac{z^n}{n! }.$$
The triangle inequality gives us $$\ldots \le \sum_{n=N+1}^\infty\frac{|z|^n}{n! }.$$
By the distributivity law for real numbers and the properties of the absolute value for real numbers, $$\ldots =\frac{|z|^{N+1}}{(N+1)!}\left(1 + \frac{|z|}{N+2} + \frac{|z|^2}{(N+2)(N+3)} + \ldots + \frac{|z|^k}{(N+2)\cdot\ldots\cdot(N+k+1)} + \ldots \right).$$
Finally, by the rules for calculating with inequalities or real numbers we get $$\le\frac{|z|^{N+1}}{(N+1)!}\left(1 + \frac{|z|}{N+2} + \left(\frac{|z|}{(N+2)}\right)^2 + \ldots + \left(\frac{|z|}{(N+2)}\right)^k + \ldots \right).$$
If we require that $|z|\le\frac{N+2}{2}$ then we have \[\frac{|z|}{N+2}\le \frac{\frac{N+2}{2}}{N+2}=\frac 12.\]
Therefore, for $|z|\le\frac{N+2}{2}$ it follows with the formula for the infinite geometric series. $$|r_{N + 1}(z)|\le \frac{|z|^{N+1}}{(N+1)!}\left(1 + \frac{1}{2} + \frac{1}{4} + \ldots + \frac{1}{2^k} + \ldots \right)=2\frac{|z|^{N+1}}{(N+1)!}.$$
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Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983