We have \begin{align*} \dbinom{4}{0}&+\dbinom{4}{1}+\dbinom{4}{2}+\dbinom{4}{3}+\dbinom{4}{4} \\ &= \dfrac{4!\phantom{}}{0!4!\phantom{}} + \dfrac{4!\phantom{}}{1!3!\phantom{}} + \dfrac{4!\phantom{}}{2!2!\phantom{}} + \dfrac{4!\phantom{}}{3!1!\phantom{}} + \dfrac{4!\phantom{}}{4!0!\phantom{}} \\ &= \dfrac{4\cdot 3\cdot 2\cdot 1}{4\cdot 3\cdot 2\cdot 1} + \dfrac{4\cdot 3\cdot 2\cdot 1}{1\cdot 3\cdot 2\cdot 1} + \dfrac{4\cdot 3\cdot 2\cdot 1}{2\cdot 1\cdot 2\cdot 1}\\ &\qquad + \dfrac{4\cdot 3\cdot 2\cdot 1}{3\cdot 2\cdot 1\cdot 1} + \dfrac{4\cdot 3\cdot 2\cdot 1}{4\cdot 3\cdot 2\cdot 1} \\ &= 1 + 4 + \dfrac{4\cdot3}{2} + 4 + 1 \\ &= 1 + 4 + 6 + 4 + 1 \\ &= \boxed{16}.\end{align*} Notice that this is $ 2^4 $. There is an explanation for this: $$\dbinom{4}{0}+\dbinom{4}{1}+\dbinom{4}{2}+\dbinom{4}{3}+\dbinom{4}{4}$$is the sum of the number of ways to choose $ 0 $, $ 1 $, $ 2 $, $ 3 $, or $ 4 $ out of $ 4 $ objects. This accounts for every way of choosing any number of the $ 4 $ objects. There are $ 2\times 2\times 2\times 2 = 2^4 $ ways of doing so, because there are two options for each object: choose that object or don't choose it.