1088?

Note that $a_1$ , $a_2$ , $a_3$ , $a_4$ , $a_5$ are the roots of the equation $x^5 - 1088 = 0.$ Hence, the $(a_i + 4)$ 's are the roots of the equation $(x-4)^5 - 1088 = 0, \\ x^5 + \dots - 2112 = 0.$ Since the product of the roots of the equation above is given by $- (-2112)/1 = 2112$ , the desired product is $\boxed{2112}$ .

$a_{1},a_{2},a_{3}a_{4},a_{5}$ are the roots of the equation $x^{5}=1088$

By this we get that $a_{1}.a_{2}.a_{3}.a_{4}.a_{5}=1088$

The Given expression can be simplified to get $4^{5}+4^{4}(a_{1}+a_{2}+a_{3}+a_{4}+a_{5})+.............+a_{1}.a_{2}.a_{3}.a_{4}.a_{5}$

All terms will become zero acc. to $x^{5}=1088$ except first and the last term.

So the result obtained will be $=4^{5}+1088$

$\large{=2112}$

Let us define

$f(x)=x^5-1088\quad \boxed{1}$ .

Then $f(a_1)=f(a_2)=f(a_3)=f(a_4)=f(a_5)=0$ , since $a_1, a_2,a_3,a_4,a_5$ are the roots of the equation $x^5-1088=0$ .

However we can also factorise the polinomial $x^5-1088$ in terms of its roots as

$f(x)=(x-a_1)(x-a_2)(x-a_3)(x-a_4)(x-a_5)\quad \boxed{2}$ .

Using $\boxed{2}$ , we now write

$f(-4)=-(4+a_1)(4+a_2)(4+a_3)(4+a_4)(4+a_5),$

Alternatively, using the $\boxed{1}$ ,

$f(-4)=(-4)^5-1088=-2112$ .

And that's the end of the story.

As a side note, we can write $1088=1024+64=2^{10}+2^6=2^{10} (1+2^{-4})$ . Since for the sake of estimation, we note that $2^{-4}\ll 1$ . Then $1088\approx 2^{10}$ , so the real fifth root of 1088 is very close to 4.