What is the proof for $F(x)=\int_{a}^{x} e^{t^2}dt$ not being elementary?

Let's use the result of Liouville https://math.stackexchange.com/a/163/442

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Let $f, g$ be rational functions, $g$ not constant. The indefinite integral $$ \int f(x)e^{g(x)}\;dx $$ is elementary if and only if there is a rational function $h$ so that $f=h'+hg'\;$.

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Take $f(x) = 1$, $g(x) = x^2$ there. The theorem tells us that the indefinite integral $\int e^{x^2}\,dx$ is elementary if and only if there is a rational function $h(x)$ such that $f=h'+hg'\;$; that is: $$ 1 = h'(x)+2xh(x) . \tag1$$ Now if ($1$) holds in any interval of the real line, then it holds in all of $\mathbb C$.

In the linear differential equation with analytic coefficients ($1$), every point is a regular point. So the solution has no poles. The rational function $h$ is, in fact, a polynomial. Now let us study the behavior as $x \to \infty$. I claim $h(x)$ is bounded as $x \to \infty$. Suppose not. The Laurent series near $x=\infty$ is $$ h(x) = c x^m+O(x^{m-1})\qquad\text{as }x\to\infty $$ with $c\ne 0, m \ge 1$. So $$ h'(x) = cmx^{m-1}+O(x^{m-2}) $$ and $$ h'(x)+2xh(x) = 2 c x^{m+1} +O(x^{m}) $$ But $m \ge 1$ and $2c \ne 0$, so this is not the constant $1$.

Summary: $h$ is a polynomial, $h(x)$ is bounded as $x\to\infty$. Therefore $h$ is constant. Which again contradicts ($1$).