This problem is popular among the math lore

Let $f(x) = \lfloor x \lfloor x \lfloor x \lfloor x \rfloor\rfloor\rfloor\rfloor$ , so we want to solve for minimum value of $x$ such that $f(x) = 88$ . It's easy to see that $f(x)$ is a non-decreasing function.

Because $f(3) < 88 < f(4)$ . $x$ be must be in the interval of $(3,4)$ , so $\lfloor x \rfloor = 3$ . Thus $f(x) = \lfloor x \lfloor x \lfloor3x\rfloor\rfloor\rfloor$ .

Now suppose that $x \geq 3 + \frac13$ , then $f(x) \geq f\left(3 + \frac13 \right) = \left \lfloor \left(3 + \frac13\right) \left \lfloor \left(3 + \frac13\right) \cdot 10 \right \rfloor \right \rfloor = 110 > 88$ . So $3 < x< 3 + \frac13$ . Therefore $\lfloor 3x \rfloor = 9$ , and $f(x) = \lfloor x \lfloor 9x \rfloor \rfloor$ .

Again, suppose that $x \geq 3 + \frac29$ , then $f(x) \geq f\left(3 + \frac29 \right) = \left \lfloor \left(3 + \frac29\right) \left \lfloor \left(3 + \frac29\right) \cdot 9 \right \rfloor \right \rfloor = 93 > 88$ . So $3 < x< 3 + \frac29$ , this means that $\lfloor 9x \rfloor$ equals to 27 or 28.

Suppose $\lfloor 9x \rfloor = 27$ , then $f(x) = \lfloor 27x \rfloor$ , then $\left \lfloor 27(3 + y) \right \rfloor = 88$ for some real $0 < y < \frac19$ . With $\left \lfloor 27(3 + y) \right \rfloor = 88$ , $81 + \lfloor 27y \rfloor = 88 \Rightarrow \lfloor 27y \rfloor = 7$ which does not have a solution because $\lfloor 27y \rfloor < 27\cdot \frac19 = 3 < 7$ .

Therefore $\lfloor 9x \rfloor = 28$ only, and $f(x) = \lfloor 28x \rfloor$ .

Finally, we want to find the minimum value of $x$ such that $\lfloor 28x \rfloor = 88$ . Then $\left \lfloor 28(3 + y') \right \rfloor = 88$ for some real $\frac19 \leq y' < \frac29$ .

$84+ \lfloor 28y' \rfloor = 88 \Rightarrow \lfloor 28y' \rfloor = 4 \Rightarrow \min(y') = \frac17$ . Hence the minimum value of $x$ is $3 + \min(y') = 3 + \frac17 = \boxed{\frac{22}7}$ .

The dessert gave it away.