Proving that real part of $\exp$ is positive and increasing

Question

Let's define $\exp(z)$ in the following form: $$\exp(z) = \sum \limits_{n=0}^{\infty} \frac{z^n}{n!}, z \in \mathbb{C}.$$ We are to show that the real part of the function defined above is increasing and positive $\forall_{x \in \mathbb{R}}$.
I found some information linked to $\exp$ and my problem here but I don't understand the proof and I don't know if it's enough.

Answer 1

We have a function $f: \mathbb R \to \mathbb R$ such that $f(0)=1$ and $f>1$ on $(0,\infty).$ We also know $f(x+y)= f(x)f(y)$ for all $x,y\in \mathbb R.$ Thus for any real $x,$ $f(x-x)=f(0)=1=f(x)f(-x),$ which implies $f(-x)=1/f(x).$ From this it follows that $f>0$ on $(-\infty,0).$ Hence $f>0$ everywhere.

Finally, suppose $x<y.$ Then $0<y-x,$ hence

$$1=f(0) < f(y-x) = f(y)f(-x)= f(y)/f(x).$$

This implies $f(x) < f(y)$ and we're done.

Answer 2

$\exp(0) = 1$

If $x>0$ then $\exp(x)>1$

$\exp(x) - 1 = \sum_\limits{n=1}^{\infty} \frac {x^n}{n!}$

every term on the right is positive, so the sum must be positive.

$\exp(x+h) = \exp(x)\exp(h)$

$\exp(x-x) = \exp(x)\exp(-x) = exp(0) = 1\\ \exp(-x) = \frac {1}{\exp(x)}$

For all real $x, \exp(x)>0$

$\exp(x)$ is increasing if for all $h>0, \exp(x+h) > \exp(x)$

$\exp(x)\exp(h)>\exp(x)\\ \exp(x)(\exp(h)-1)>0$

Since both factors are positive, it is clearly true.