Using Cauchy's generalised integral formula, show that if $f$ is an entire function and $|f(z)|\leq |z|+1$, then $f^{(k)}(0)=0$.
Hence deduce that there are complex constants $a,b$ such that $f(z)=az+b$ where $|a|\leq 1$ and $|b|\leq 1$.
I have shown that $f^{(k)}(0)=0$, as for $$f^{(k)}(0)=\frac{n!}{2\pi i}\int_\Gamma \frac{f(z)}{z^{n+1}} \ dz\implies|f^{(k)}(0)|\leq\frac{n!}{R^{n-1}}+\frac{n!}{R^n}.$$ So for $n\geq 2$, $f^{(k)}(0)=0$ as $R\rightarrow\infty$, by the ML lemma. Hence, by Cauchy's generalised integral formula, $$f(z)=f(0)+f'(0)z\implies a=f'(0), \ b=f(0).$$
My question is, how do we know that $|a|\leq 1$ and $|b|\leq 1$?
attempt
From $|f(z)|\leq |z|+1\implies |f(0)|\leq |0|+1=1$, hence $|b|\leq 1$. Next, consider \begin{align} f(z)&=f(0)+f'(0)z \\ f'(0)z&=f(z)-f(0) \\ |f'(0)z|&\leq |f(z)|-|f(0)| \ \ \text{(triangle inequality)} \\ |f'(0)z|&\leq |z|+1-1 \\ |f'(0)|&\leq \frac{|z|}{|z|}=1, \end{align} hence $|a|\leq 1$.