Here is a proof that doesn't seem to employ induction but uses the uniqueness of the binary representation of the non-negative integers (which may use induction).
Let $0\le m\lt2^n$ then there is only one way to get $x^m$ from the product $ (1+x)(1+x^2)(1+x^4)\dots(1+x^{2^{n-1}}) $ that is by choosing $x^{2^k}$ in the factors where bit $k$ is $1$ in the binary representation of $m$ and choosing $1$ in the factors where bit $k$ is $0$. All coefficients in the factors are $1$ so the coefficient of $x^m$ in the product is $1$.
Therefore, $ (1+x)(1+x^2)(1+x^4)\dots(1+x^{2^{n-1}})=1+x+x^2+x^3+\dots+x^{2^n-1} $