Closed form of function from recursive definition

The binomial coefficient-weighted sum on the right-hand side suggests using exponential generating functions. Define $f(0)=0$ and rewrite the given equation as $$ \sum_{j=0}^k \binom kj f(j) = (2^k-1)^n = \sum_{i=0}^n \binom ni (2^k)^i (-1)^{n-i} $$ (assuming $n$ is a nonnegative integer). Multiplying both sides by $x^k/k!$ and summing over $k$ yields the exponential generating function identity \begin{align*} \sum_{k=0}^\infty \frac{x^k}{k!} \sum_{j=0}^k \binom kj f(j) &= \sum_{k=0}^\infty \frac{x^k}{k!} \sum_{i=0}^n \binom ni (2^k)^i (-1)^{n-i} \\ &= \sum_{i=0}^n (-1)^{n-i} \binom ni \sum_{k=0}^\infty \frac{(2^ix)^k}{k!} = \sum_{i=0}^n (-1)^{n-i} \binom ni e^{2^ix}. \end{align*} For the left-hand side, we use the fundamental convolution identity $$ \biggl( \sum_{k=0}^\infty a_k \frac{x^k}{k!} \biggr) \biggl( \sum_{k=0}^\infty b_k \frac{x^k}{k!} \biggr) = \sum_{k=0}^\infty \biggl( \sum_{j=0}^k \binom kj a_j b_{k-j} \biggr) \frac{x^k}{k!} $$ with $a_j = f(j)$ and $b_j=1$, so that $$ \sum_{k=0}^\infty \frac{x^k}{k!} \sum_{j=0}^k \binom kj f(j) = \biggl( \sum_{k=0}^\infty f(k) \frac{x^k}{k!} \biggr) \biggl( \sum_{k=0}^\infty \frac{x^k}{k!} \biggr) = e^x \sum_{k=0}^\infty f(k) \frac{x^k}{k!}. $$ We conclude that $$ \sum_{k=0}^\infty f(k) \frac{x^k}{k!} = e^{-x} \sum_{i=0}^n (-1)^{n-i} \binom ni e^{2^ix} = \sum_{i=0}^n (-1)^{n-i} \binom ni e^{(2^i-1)x}. $$ Comparing Maclaurin coefficients then gives $$ f(k) = \sum_{i=0}^n (-1)^{n-i} \binom ni (2^i-1)^k. $$ Note that this is consistent with the formulas in Raymond Manzoni's answer.