Proving $\int_0^\infty e^{-x^p} dx$ converges or not

No, you should separate into cases involving $p \leq 0$ and $p > 0$.

If $p \leq 0$, then ${-x^p} \to l$ as $x \to \infty$ for some finite $l$, so $e^{-x^p} \to e^{l}$ as $x \to \infty$. There's no way the integral can exist in this case : if the integrand coverges to a non-zero value, the integral can't converge.

If $p >0$, then we will show that for every $n$, the integral $\int_{0}^\infty e^{-\sqrt[n]{x}} dx$ exists. By monotonicity, the integral will exist for all $p > 0$, since for every $p$ there is $N$, $p > \frac 1N$, and then you can use the domination.

So make a change of variable : $t = \sqrt[n]x$, then $t^n = x$ so $nt^{n-1}dt = dx$.

From here , we get $n\int_{0}^\infty t^{n-1}e^{-t} dt$. Now, prove for yourself the following relation using integration by parts : $$ \forall k \geq 1 \quad\int_{0}^\infty t^{k}e^{-t}dt = k\int_{0}^\infty t^{k-1}e^{-t}dt $$

From here, conclude by induction that in fact, $\int_{0}^\infty e^{-\sqrt[n]{x}}dx = 1 \times 2 \times ... \times n = n! < \infty$.

Assuming $p>0$, let $$x^p=t \implies x=t^{\frac{1}{p}}\implies dx=\frac{1}{p}t^{\frac{1}{p}-1}$$

$$\int e^{-x^p} dx=\frac{1}{p}\int e^{-t}\,t^{\frac{1}{p}-1}\,dt=-\frac{1}{p}\,\Gamma \left(\frac{1}{p},t\right)$$ $$\int_0^{+\infty} e^{-x^p} dx=\frac{1}{p}\int_0^{+\infty} e^{-t}\,t^{\frac{1}{p}-1}\,dt=\frac{1}{p}\,\Gamma \left(\frac{1}{p},0\right)\quad > \quad \forall p > 0$$