Proving that product of roots does not exceed $\frac{1}{2^n}$ if $|f(0)| = f(1)$ and the roots lie between 0 to 1.

Problem: Let $$a_0, a_1, \ldots, a_{n-1}$$ be real numbers where $$n \geq 1$$ and let $$f(x) = x^n + a_{n-1} x^{n-1} + a_{n-2} x^{n-2} + \ldots + a_0$$ be such that $$|f(0)| = f(1)$$ and each root of $$f(x) = 0$$ is real and lies between $$0$$ to $$1$$. Prove that the product of the roots of this equation does not exceed $$\frac{1}{2^n}$$

This problem was brought to my attention through a friend. I could only obtain weak bounds by using AM-GM along with Vieta's lemma, which is nowhere close to the bound required. I thought of some relation with sum of binomial coefficients of $$(1+x)^n$$, where $$x = 1$$, but that also got me nowhere.

Any help with this problem would be appreciated.

Let $$f(x) = (x-\lambda_1)(x-\lambda_2)\dots (x-\lambda_n)$$, where $$0 \leq \lambda_1 \leq \dots \leq \lambda_n \leq 1$$ are the roots of $$f$$.
The product of the roots is $$\pi_n := \lambda_1 \dots \lambda_n = |f(0)|$$, and we also have $$|f(0)| = f(1) = (1-\lambda_1) \dots (1-\lambda_n)$$.
Then we have: $$\pi_n^2 = |f(0)|f(1) = \lambda_1(1-\lambda_1)\dots \lambda_n(1-\lambda_n) \leq \frac{1}{4^n}$$
Where the upper bound comes from the fact that $$x(1-x) \leq 1/4$$ for $$x \in [0,1]$$.
Therefore we deduce that $$\pi_n \leq 1/2^n$$.