A question about complex exponential The famous power series $1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\ldots$ converges normally in the whole complex plane $\mathbb C$ and mathematicians chose to call
$$\exp(z):=\sum_{n=0}^{\infty} \frac{z^n}{n!}$$
now my question is:
Why, they where sure that the above limit wasn't a rational function? In a nutshell, who does ensure that there aren't two polynomials $p$ and $q$ such that $\frac{p}{q}=\sum_{n=0}^{\infty} \frac{z^n}{n!}$ ?
 A: There are several possible ways to eliminate the possibility of a rational function, for example:
It is not hard to prove that $\exp(z)$ tends to infinity faster than any power of $x$.
Also, looking at the function in the complex plane, it is periodic, and it is not hard to prove that a rational function can never be periodic.
A: A rational function has a limit as $z\to\infty$ (either a complex number or $\infty$). On the other hand, $\lim_{z\to\infty}e^z$ does not exist.
A different reasoning. Since $e^z$ has no singularities other than $\infty$, if it were rational, it should be a (non-constant) polynomial. But a nonconstant polynomial vanishes at some point, while $e^z\ne0$ for any complex number $z$.
A: $$\exp(z):=\sum_{n=0}^{\infty} \frac{z^n}{n!}$$
Let's assume that we have 2 polynomials and we express $e^x=\frac{P(x)}{Q(x)}$
The polynomials are
$$P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0$$
$$Q(x)=b_mx^m+b_{m-1}x^{m-1}+\cdots+b_0$$
$(e^x)'=e^x=\left(\frac{P(x)}{Q(x)}\right)'$
We assumed $e^x=\frac{P(x)}{Q(x)}$
Thus,
$$e^x=\frac{P(x)}{Q(x)}=\left(\frac{P(x)}{Q(x)}\right)'$$
$$\frac{P(x)}{Q(x)}=\left(\frac{P(x)}{Q(x)}\right)'=\left(\frac{P'(x)Q(x)- P(x)Q'(x)}{Q^2(x)}\right)$$
$$\frac{P(x)}{Q(x)}=\frac{P'(x)Q(x)- P(x)Q'(x)}{Q^2(x)}$$
$$P(x)Q(x)=P'(x)Q(x)- P(x)Q'(x)$$
$$P(x)(Q(x)+Q'(x))=P'(x)Q(x)$$
$$P(x)(Q(x)+Q'(x))=(a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0)(b_mx^m+(mb_m+b_{m-1})x^{m-1}+\cdots)=a_nb_mx^{n+m}+\cdots$$
$$P'(x)Q(x)=(na_nx^{n-1}+(n-1)a_{n-1}x^{n-2}+\cdots+a_1)(b_mx^m+b_{m-1}x^{m-1}+\cdots+b_0)=na_nb_mx^{n+m-1}+\cdots$$
As we see $P(x)(Q(x)+Q'(x))=P'(x)Q(x)$ cannot be equal because their degrees are not equal. The degree of polynomial $P(x)(Q(x)+Q'(x))$ is $m+n$, The degree of other polynomial $P'(x)Q(x)$ is $m+n-1$ . Thus it is impossible to find $e^x=\frac{P(x)}{Q(x)}$ where  $P(x)$  and $Q(x)$ are polynomials.
A: Another argument that can easily be checked with power series: 
$$\left(\sum_{n=0}^\infty\frac{z^n}{n!}\right)'=\sum_{n=0}^\infty\frac{z^n}{n!}$$  which cannot be true for no non-zero rational functions.
A: If a rational function of $z$ has a limit as $z\to-\infty$, then it has the same limit as $z\to\infty$.  If you can show that this function approaches $0$ as $z\to-\infty$ and approaches $\infty$ as $z\to\infty$, then you've shown it doesn't behave at all like a rational function.
