The given integral is:

$\int_0^\infty x^0 e^{-x} \, dx$

First, observe that $x^0 = 1$ for all $x$, so the integral simplifies to:

$\int_0^\infty e^{-x} \, dx$

This is a standard integral, which can be computed as follows:

$\int_0^\infty e^{-x} \, dx = \left[-e^{-x}\right]_0^\infty$

At $x = \infty$, $e^{-\infty} = 0$, and at $x = 0$, $e^0 = 1$, so:

$\left[-e^{-x}\right]_0^\infty = 0 - (-1) = 1$

Thus, the value of the integral is:

$\boxed{1}$

Would you like further details or explanations on any step?

Here are 5 related questions for further practice:

1. What is $\int_0^\infty x^n e^{-x} \, dx$ for any positive integer $n$?
2. How does the Gamma function generalize this type of integral?
3. What is the value of $\int_0^\infty e^{-ax} \, dx$ for any positive constant $a$?
4. How would you compute $\int_0^1 e^{-x} \, dx$?
5. What happens if the upper bound of the integral changes to a finite number instead of infinity?

Tip: For integrals involving $e^{-x}$ and powers of $x$, the Gamma function $\Gamma(n)$ can be a helpful tool to generalize the solution.