Use the power series definition of $e^z$ to prove that $|\frac{t^h-1}{h}|\leq t^{|h|}|\ln t|$ Use the power series definition of $e^z$ to prove that for any $h\in \mathbb{C}$ and $t>0$, we have 
$$\left|\frac{t^h-1}{h} \right|\leq t^{|h|}|\ln t|$$ 
and furthermore, 
$$\left|\frac{t^h-1}{h}-\ln t \right|\leq |h|t^{|h|}\ln^{2} t$$
I first tried to approach it first by using the fact that $t^h=e^{h\ln t}$ to get 
$$\left|\frac{t^h-1}{h} \right| = \left|\frac{e^{h\ln t}-1}{h} \right|$$
Since the power series expansion of $e^z$ is $\sum_{0}^{\infty}\frac{z^n}{n!}$, I then substituted that to get 
$$\left|\frac{e^{h\ln t}-1}{h} \right| = \left|\frac{(\sum_{0}^{\infty}\frac{z^n}{n!})^{\ln t}-1}{h} \right|$$
But then, I don't know how I should be approaching this problem. I feel like I'm no where near the right track.
 A: Starting from the comment :
$$\left|\frac{\sum_{k=1}^{\infty}\frac{(h\ln t)^{k}}{k!}}{h}\right| \leq
\frac{\sum_{k=2}^{\infty}|\frac{(h\ln t)^{k}}{k!}|}{|h|}\\
\leq \frac{\sum_{k=2}^{\infty}\frac{|h|^k|\ln t|^{k}}{k!}}{|h|}\\
\leq \sum_{k=2}^{\infty}\frac{|h|^{k-1}|\ln t|^{k}}{k!}\\
\leq |\ln t|\sum_{k=2}^{\infty}\frac{|h|^{k-1}|\ln t|^{k-1}}{k!}$$
Then you notice that :
$$\frac{|h|^{k-1}|\ln t|^{k-1}}{k!}\leq \frac{|h|^{k-1}|\ln t|^{k-1}}{(k-1)!}$$
So we have :
$$\left|\frac{\sum_{k=1}^{\infty}\frac{(h\ln t)^{k}}{k!}}{h}\right| \leq |\ln t|\sum_{k=2}^{\infty}\frac{|h|^{k-1}|\ln t|^{k-1}}{(k-1)!}\\
\leq |\ln t|\sum_{k=1}^{\infty}\frac{|h|^{k}|\ln t|^{k}}{k!}$$
You notice that :
$$\sum_{k=1}^{\infty}\frac{|h|^{k}|\ln t|^{k}}{k!}=\sum_{k=1}^{\infty}\frac{|h\ln t|^{k}}{k!}=\sum_{k=0}^{\infty}\frac{|h\ln t|^{k}}{k!}-1=|t|^{|h|}-1=t^{|h|}-1$$
The last equality is true because $t>0$.
Finally :
$$\left|\frac{\sum_{k=1}^{\infty}\frac{(h\ln t)^{k}}{k!}}{h}\right| \leq |\ln t|(t^{|h|}-1)\leq |\ln t|t^{|h|}$$
