We can do it by brute force, expanding all 108 product terms and combining like terms carefully. Many of the terms can be eliminated immediately.
In hindsight or by luck or by sharp observation that D and E appear most frequently, we can proceed as follows.
$(\overline{A}+\overline{B}+E)(\overline{A}+\overline{C}+D)(C+D+\overline{E})(\overline{B}+D)(A+E)$
Let $D=1$ and $E = 1$. The formula becomes 11111 = 1
Let $D=1$ and $E = 0$. The formula becomes $(\overline{A}+\overline{B})1 1 1 A = \overline{B}A$
Let $D=0$ and $E =1$. The formula becomes $1(\overline{A}+\overline{C}) C \overline{B} 1 = \overline{A}C\overline{B}$
Let $D=0$ and $E = 0$. The formula becomes $(\overline{A}+\overline{B})(\overline{A}+\overline{C})1 \overline{B} A = \overline{C}\overline{B}A$
So the original formula is
$$DE + (D\overline{E}A + \overline{D}(E\overline{A}C + \overline{E}\overline{C}A))\overline{B}$$
(Computationally, it is debatable which formula is simpler.)