Consider the polynomial \begin{align*} p(x) &= \frac{a^5 (x - b)(x - c)(x - d)(x - e)}{(a - b)(a - c)(a - d)(a - e)} + \frac{b^5 (x - a)(x - c)(x - d)(x - e)}{(b - a)(b - c)(b - d)(b - e)} \\ &\quad + \frac{c^5 (x - a)(x - b)(x - d)(x - e)}{(c - a)(c - b)(c - d)(c - e)} + \frac{d^5 (x - a)(x - b)(x - c)(x - e)}{(d - a)(d - b)(d - c)(d - e)} \\ &\quad + \frac{e^5 (x - a)(x - b)(x - c)(x - d)}{(e - a)(e - b)(e - c)(e - d)}.\end{align*}Note that $ p(x) $ is a polynomial of degree at most 4. Also, $ p(a) = a^5, $ $ p(b) = b^5, $ $ p(c) = c^5, $ $ p(d) = d^5, $ and $ p(e) = e^5 $. This might lead us to conclude that $ p(x) = x^5, $ but as we just observed, $ p(x) $ is a polynomial of degree 4. So, consider the polynomial \[q(x) = x^5 - p(x).\]The polynomial $ q(x) $ becomes 0 at $ x = a, $ $ b, $ $ c, $ $ d, $ and $ e $. Therefore, \[q(x) = x^5 - p(x) = (x - a)(x - b)(x - c)(x - d)(x - e) r(x)\]for some polynomial $ r(x) $. Since $ p(x) $ is a polynomial of degree at most 4, $ q(x) = x^5 - p(x) $ is a polynomial of degree 5. Furthermore, the leading coefficient is 1. Therefore, $ r(x) = 1, $ and \[q(x) = x^5 - p(x) = (x - a)(x - b)(x - c)(x - d)(x - e).\]Then \[p(x) = x^5 - (x - a)(x - b)(x - c)(x - d)(x - e),\]which expands as \[p(x) = (a + b + c + d + e) x^4 + \dotsb.\]This is important, because the expression given in the problem is the coefficient of $ x^4 $ in $ p(x) $. Hence, the expression given in the problem is equal to $ a + b + c + d + e $. By Vieta's formulas, this is $ \boxed{-7} $.