 Satisfying-assignment count for nested implications

For distinct propositional variables, eliminate each right-associated logical implication to obtain $\neg A_n\vee \cdots \vee \neg A_2\vee A_1$. Exactly one Boolean valuation falsifies all these disjuncts: $A_1=0$ and $A_2=\cdots =A_n=1$. Of the $2^n$ assignments, the other $2^n-1$ are satisfying assignments. Left-associated implication chains have a different meaning and need not have the same count.