A polynomial $f(x) = x^n + a_{n-1}x^{n-1} \cdots + a_1x + a_0,$ with integer coefficients $a_i,$ has roots at $1, 2, 4, \ldots, 2^{n-1}.$ 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 $x-x_0.$ How does this help you answer the first hint?

… more specifically, $x-x_0$ is a factor of $f(x),$ i.e. $f(x)=(x-x_0)g(x)$ where $g(x)$ is a polynomial of degree $n-1.$ 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 $f(x)=\prod_{i=0}^{n-1}(x - 2^i).$ Thus, \begin{aligned} f(0)&=\prod_{i=0}^{n-1}-2^i \\&= \prod_{i=0}^{n-1}(-1)\cdot2^i \\&= (-1)^n\;2^{\sum_0^{n-1}i} \\&= (-1)^n\;2^{n(n-1)/2}. \end{aligned} There are no duplicate roots since $f$ is of degree $n$ and we were told it has $n$ distinct roots.