Proving a function is NOT of exponential order

Question

Prove that the function $e^{x^2}$ is not of exponential order. For contradiction, assume that $e^{x^2}$ is of exponential order. Then, by definition, $e^{x^2} \leq Me^{cx}$ for some $M,c$ . But this implies that $M\geq 0$, since $e^{x^2}\geq 0$, so taking the logarithm will be defined. I have: $e^{x^2-cx} \leq M$ But then by taking the natural logarithm I get: $x^2-cx\leq logM$. This is essentially where I am stuck on reaching the contradiction. I feel I am done because it is obvious that a quadratic of this form is not bounded above by a fixed constant. However, I feel that this is not the best contradiction for this problem. Any advice starting from my last step would be appreciated. (Or if I made an error reaching my last step please tell me). Note, assume that we are working in the field of real numbers.