1. **State the problem:** Evaluate the integral $$\int_{-\infty}^{0} e^{x} \sqrt{1+e^{x}} \, dx$$. 2. **Substitution:** Let $$t = \sqrt{1+e^{x}}$$, so $$t^{2} = 1 + e^{x}$$. 3. Differentiate both sides: $$2t \, dt = e^{x} \, dx$$, hence $$e^{x} \, dx = 2t \, dt$$. 4. **Change limits:** When $$x \to -\infty$$, $$e^{x} \to 0$$, so $$t = \sqrt{1+0} = 1$$. When $$x = 0$$, $$e^{0} = 1$$, so $$t = \sqrt{1+1} = \sqrt{2}$$. 5. **Rewrite the integral:** $$\int_{-\infty}^{0} e^{x} \sqrt{1+e^{x}} \, dx = \int_{t=1}^{\sqrt{2}} t \cdot (2t \, dt) = 2 \int_{1}^{\sqrt{2}} t^{2} \, dt$$. 6. **Integrate:** $$2 \int_{1}^{\sqrt{2}} t^{2} \, dt = 2 \left[ \frac{t^{3}}{3} \right]_{1}^{\sqrt{2}} = \frac{2}{3} \left( (\sqrt{2})^{3} - 1^{3} \right)$$. 7. Simplify: $$(\sqrt{2})^{3} = (2^{1/2})^{3} = 2^{3/2} = 2 \sqrt{2}$$. So the value is: $$\frac{2}{3} (2 \sqrt{2} - 1)$$. **Final answer:** $$\boxed{\frac{2}{3} (2 \sqrt{2} - 1)}$$