Closed form of function from recursive definition

Define $f(x)$ for positive integer $x$ as $$f(x) = (2^x - 1)^n - \sum_{j = 1}^{x - 1}\binom{x}{j} \cdot f(j)$$where $n$ is some constant and $f(1) = 1$. I want to determine $f(x)$ explicitly. Here is what I was able to do: $$f(x) - f(x - 1) = (2^x - 1)^n - (2^{x-1}- 1)^n - \sum_{j = 1}^{x - 2}\binom{x - 1}{j - 1}\cdot f(j) - \binom{x}{x - 1}\cdot f(x - 1)$$ I tried evaluating the telescoping series, but it just became extremely complicated and unhelpful. Any ideas on how else I can find $f(x)$ in closed form?

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.

Conjectured results for the first values of $n$ : \begin{array} {cc} n& f(x)\\ 1&1\\ 2&3^x-2\\ 3&7^x-3\cdot 3^x+3\\ 4&15^x-4\cdot 7^x+6\cdot 3^x-4\\ 5&31^x-5\cdot 15^x+10\cdot 7^x-10\cdot 3^x+5\\ \end{array}

corresponding to Greg Martin's solution!