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here is one proof that I know but I am not totally sure if it is acceptable-

exponential functions are exponential: no matter how many times you differentiate them e.g-

f(x)=e^x
first derivative f`(x)= e^x
2nd derivative f``(x)= e^x
3rd derivative f```(x)=e^x

and so on.

now if you differentiate a polynomial function- let's say,

f(x)= x^5
1st derivative f`(x)= 5x^4
2nd de3rivative f``(x)= 20x^3
3rd derivatives f```(x)=60x^2
4th derivative f````(x)=120x
5th derivative f`````(x)= 0

like this every polynomial finally gets differentiated to zero or a constant . this proves that the polynomials are not exponential.

                        **is my proof ok**  

I want more alternate proofs and a brief explanation about this one.

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  • $\begingroup$ Your proof is fine if we just want to prove that $p(x)$ is not $e^x$. But how do we know that $p(x)=e^{f(x)}$ has no solution whatever function $f$ is? For example, if $p(x)=x^2+1$, $e^{\ln(p(x))}=p(x)$ which is a polynomial. $\endgroup$ – Bernard Massé May 23 '20 at 18:17
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Your proof is correct. You can also say that $\lim_{x\to-\infty}e^x=0$, whereas you have$$\lim_{x\to-\infty}P(x)=\pm\infty$$if $P$ is a non-constant polynomial function. And, clearly, the exponential function is not constant.

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Suppose $e^x=P(x)$, where $P$ is a polynomial of degree $n$. Note first that $n\gt0$, since $e^x$ is nonconstant. It follows that $P(2x)$ and $(P(x))^2$ are polynomials of different degrees, namely $n$ and $2n$. But $P(2x)=e^{2x}=(e^x)^2=(P(x))^2$ says they are of the same degree, which is a contradiction. So $e^x$ is not equal to any polynomial.

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