Show that $\displaystyle\lim_{x\rightarrow\infty}[e^x-(a_0+a_1x+a_2x^2+\dots+a_Nx^N)]=\infty$

Assume that N is given and that $p_N$ is not the zero polynomial. I want to show that:

$\displaystyle\lim_{x\rightarrow\infty}[e^x-(a_0+a_1x+a_2x^2+\dots+a_Nx^N)]=\infty$.

I am not able to do this direct and easy with l'Hôpital, are you able to see how to do it?

I am able to do it indirectly with l'Hôpital, can you please check if it is a valid proof logically?

proof:

From l'Hôpital we do know that:

$\displaystyle\lim_{x\rightarrow\infty}\left[\frac{e^x}{a_0+ax+\dots+a_Nx^N}\right]=\infty$,   (1)

Since the denominator clearly is not zero we have that:

$\dfrac{\lim\limits_{x\rightarrow\infty} e^x}{\lim\limits_{x\rightarrow\infty}(a_0+ax+\dots+a_Nx^N)}=\infty$,  (2)

Now assume for contradiction that:

$\displaystyle\lim_{x\rightarrow\infty}[e^x-(a_0+a_1x+a_2x^2+\dots+a_Nx^N)]\ne\infty$,  (3)

The limit cannot be a number $K$, because that would contradict (2). If the limit was $-\infty$, then the polynomial would after a while be larger than $e^x$ this also contradicts (2). If the limit do not exist the expression fluctuates, it means that the values the expression takes have to be bounded, this means that:

$[e^x-(a_0+a_1x+a_2x^2+\dots+a_Nx^N)]\le K$, for all $x \ge 0$, but this also contradicts (2).

Do you see how much work it is? Is there any way to simplify this?, it is not elegant to check the three special cases.

$$e^x-\left(a_nx^n+a_{n-1}x^{n-1}+\ldots+a_1x+a_0)\right)=x^n\left(\frac{e^x}{x^n}-a_n-\frac{a_{n-1}}x-\ldots-\frac{a_0}{x^n}\right)$$
Apply now arithmetic of limits, taking into account that $\;\frac{e^x}{x^n}\xrightarrow[x\to\infty]{}\infty\;$