$\exp(z):=\sum_{n=0}^{\infty} \frac{z^n}{n!}$
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.