Why isn't $\frac{1}{x}$ a polynomial? Why isn't $\frac{1}{x}$ a polynomial?

Does it directly follow from definition?  As far as I know, polynomials in $F$ are expressions of the form $\sum_{i=0}^{n} a_ix^i$, where $a_i\in F$ and $x$ is a symbol.

Or is there a nicer argument involved?

Footnote: $F$ is a field of characteristic zero.
If you want a more formal "proof", you can suppose for contradiction that $1/x$ is in fact equal to some expression $a_k x^n$ of the form $ \sum_{k=0}^n a_k x^n $:

$$ \frac{1}{x} = a_0 + a_1 x + \cdots + a_n x^n$$

multiplying through by $x$ gives:

$$ 1 = a_0 x + a_1 x^2 + \cdots + a_n x^{n+1}$$

Setting $x=0$ gives:
$$ 1 = a_0 \cdot 0 + \cdots + a_n \cdot 0^{n+1} = 0$$

a contradiction, so we must have that $1/x$ is not a polynomial.
You can first use intuition from polynomials where coefficients are from the real numbers.  Being every where continuous the image of $[-1,1]$ is a compact set, which is bounded for a polynomial. But the image of the same set under  $f(x)=  \frac1x$ is unbounded. So we can see that in reals, $1/x$ is not a polynomial.

Now take any field of characteristic 0. One can see that if $\alpha$ is any root of $g(x)$ then it has to be a root of the product  polynomial $f(x)g(x)$.
 Now assuming $f(x)=\frac1x$ is a polynomial multiply with the second polynomial $g(x)=x(x-1)$. Their product is the  polynomial $(x-1)$, this product has just 1 root namely $1$ whereas $g(x)$ has two roots, 0 and 1. This contradiction should settle the issue.