Points $E$ and $F$ are located on square $ABCD$ so that $\triangle BEF$ is equilateral. What is the ratio of the area of $\triangle DEF$ to that of $\triangle ABE$?

[asy]
pair A,B,C,D,I,F;
A=(0,0);
B=(10,0);
C=(10,10);
D=(0,10);
F=(7.4,10);
I=(0,2.6);
draw(B--I--F--cycle,linewidth(0.7));
draw(A--B--C--D--cycle,linewidth(0.7));
label("$A$",A,S);
label("$B$",B,S);
label("$C$",C,N);
label("$D$",D,N);
label("$E$",I,W);
label("$F$",F,N);
[/asy]
First, assume that $AB=1$, and let $ED = DF = x$.  Then, we have $[DEF] = \frac{x^2}{2}$ and $[ABE] = \frac{(AE)(AB)}{2} = \frac{(1-x)(1)}{2}$, so  \[\frac{[DEF]}{[ABE]} = \frac{x^2}{1-x} .\] By the Pythagorean Theorem applied to $\triangle DEF$, we have  \[EF^2 = DE^2 + DF^2 = 2x^2.\] Applying the Pythagorean Theorem to $\triangle AEB$, we have \[EB^2 = AB^2 + AE^2 = 1 + (1-x)^2 = 2 - 2x + x^2.\] Since $\triangle EFB$ is equilateral, we have $EF = EB$, so \[2x^2 = 2-2x + x^2,\]  or $x^2 = 2-2x= 2(1-x)$. Hence the desired ratio of the areas is \[\frac{[DEF]}{[ABE]} = \frac{x^2}{1-x} = \boxed{2}.\]