Given that $$ \mathcal{B}_{t}(\boldsymbol{z})=\sum_{\mathrm{k} \ge 0}(tk)^{\underline{k-1}}\frac{z^k}{k!}. $$ I should show that $$ \mathcal{B}_{t}(z)^{1-t}-\mathcal{B}_{t}(z)^{-t}=z. $$
I don't really know where to start ...
Note: $$x^{\underline{k}}=\frac{x!}{(x-k)!}.$$