$f(x) = x^n + a_{n1}x^{n1} \cdots + a_1x + a_0,$
with integer coefficients $a_i,$
has roots at $1, 2, 4, \ldots, 2^{n1}.$
What possible values can $f(0)$
take?

How else can you write a polynomial when you know all of its roots?

… specifically, a polynomial that has a root at
$x_0$
is divisible by the binomial$xx_0.$
How does this help you answer the first hint? 
… more specifically,
$xx_0$
is a factor of$f(x),$
i.e.$f(x)=(xx_0)g(x)$
where$g(x)$
is a polynomial of degree$n1.$
How does this help you answer the first hint? 
Finding
$f(0)$
is the same as evaluating this new expression at$0.$
What do you obtain? 
What manipulations can you perform on exponents when the bases are equal?

Specifically, try writing the product as a single number raised to a power, where the latter is an expression.

Looking at the expression (sum) in the power, do you notice a familiar relationship between its terms?

Seeing as the coefficient of
$x^n$
is 1, we can write$f(x)=\prod_{i=0}^{n1}(x  2^i).$
Thus,$$ \begin{aligned} f(0)&=\prod_{i=0}^{n1}2^i \\&= \prod_{i=0}^{n1}(1)\cdot2^i \\&= (1)^n\;2^{\sum_0^{n1}i} \\&= (1)^n\;2^{n(n1)/2}. \end{aligned} $$
There are no duplicate roots since$f$
is of degree$n$
and we were told it has$n$
distinct roots.