The Maclaurin series is a special case of the Taylor series expansion centered at $x = 0$. For the function $f(x) = e^x$, the Maclaurin series is derived from the function's derivatives evaluated at $x = 0$.

The general form of the Maclaurin series for a function $f(x)$ is:

$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f^{(3)}(0)}{3!}x^3 + \cdots + \frac{f^{(n)}(0)}{n!}x^n + \cdots$

For $f(x) = e^x$, all derivatives of $e^x$ are $e^x$, and since $e^0 = 1$, we have:

$f^{(n)}(0) = 1 \quad \text{for all} \, n.$

Thus, the Maclaurin series for $e^x$ is:

$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots + \frac{x^n}{n!} + \cdots$

This is an infinite series that represents $e^x$ as a sum of powers of $x$.

Do you have any questions, or would you like more details?

Here are 5 related questions to deepen your understanding:

1. How would the Maclaurin series change if the function were $e^{2x}$?
2. How can you use the Maclaurin series to approximate $e^x$ for small values of $x$?
3. What is the general formula for the remainder (error term) in a Taylor or Maclaurin series?
4. How does the convergence of the Maclaurin series for $e^x$ compare to other functions?
5. Can you derive the Maclaurin series for $\sin(x)$ or $\cos(x)$?

Tip: The Maclaurin series is most useful for approximating functions near $x = 0$, so it's especially valuable in fields like numerical analysis and physics.