1. The problem is to find the Maclaurin series expansion of the function $e^x$ up to the 3rd degree term. 2. The Maclaurin series for a function $f(x)$ is given by: $$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots$$ 3. For $f(x) = e^x$, all derivatives are $f^{(n)}(x) = e^x$, so at $x=0$, $f^{(n)}(0) = 1$ for all $n$. 4. Substitute these values into the Maclaurin series formula up to the 3rd degree: $$e^x \approx 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{6}$$ 5. This is the Maclaurin polynomial of degree 3 for $e^x$. Final answer: $$1 + x + \frac{x^2}{2} + \frac{x^3}{6}$$