# how would you prove that polynomial functions are not exponential?

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**
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• 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. – Bernard Massé May 23 '20 at 18:17
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.
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.