Show that $\log|e^z-z|\leq c|z|$ for $|z|>R$ 
Show that $\log|e^z-z|\leq c|z|$ for $|z|>R$.

Attempt:
$\log$ is increasing thus from the triangle inequality $$\log|e^z-z|\leq \log(|e^z|+|z|)$$
But I'm not sure how to proceed. (For very small $\Re (z)$, $|z|$  is not less then or equal to $|e^z|$).
 A: Good start. Now further, $|e^z| \le e^{|z|}$, hence 
$$
\log|e^z-z|\leq \log(e^{|z|}+|z|) = |z|+ \log(1 + \frac{|z|}{e^{|z|}})
$$
Now $\frac{|z|}{e^{|z|}}$ has a maximum at $|z| =1$, so 
$$
\log|e^z-z|\leq  |z|+ \log(1 + \frac{1}{e})
$$
From here, you can use the condition $|z| > R$ to compute 
$$
R + \log(1 + \frac{1}{e}) = R \cdot c_0(R)
$$
or 
$$
c_0(R) = 1 + \frac{\log(1 + \frac{1}{e})}{R} 
$$
So for any $c \ge c_0 (R) $ and $|z| > R$, the inequality is established.$\qquad \Box$
P.S.: This might not be the tightest value for $c$.
A: A stronger inequality is true: $|e^{z}-z|=|1+\frac {z^{2}} {2!} +\frac {z^{3}} {3!}+\cdots| \leq 1+\frac {|z|^{2}} {2!} +\frac {|z|^{3}} {3!} +\cdots \leq e^{|z|}$ for all $z$ so $\log \, |e^{z}-z| \leq |z|$ for all $z$. 
A: Just a complementary note to @Andreas' answer. From
$$\log(1+x)\leq x, \forall x> -1$$
we have
$$\log\left(1+\frac{|z|}{e^{|z|}}\right)\leq \frac{|z|}{e^{|z|}}$$
leading to
$$...\leq |z|+\frac{|z|}{e^{|z|}}=|z|\left(1+\frac{1}{e^{|z|}}\right)\leq 
|z|\left(1+\frac{1}{e^{R}}\right)$$
