Since, $(p \rightarrow (q \rightarrow $ is repeating. I can consider it as a recurrence problem and write the recurrence as:
$$A_n = (p \rightarrow (q \rightarrow A_{n-2})) $$
where $A_{n-2}$ has $ \#p + \#q = n-2 $
Now, $A_n =\; \sim p \; \vee \; \sim q \; \vee A_{n-2}$
$A_n =\; \sim p \; \vee \; \sim q \; \vee \; \sim p \; \vee \; \sim q \;\vee A_{n-4} = \; (\sim p \; \vee \sim p) \vee \; (\sim q \; \vee \sim q) \vee A_{n-4} $
$A_n =\; \sim p \; \vee \; \sim q \; \vee A_{n-4} $
If we keep unrolling $A_{n-k}$ and write it as $\sim p \; \vee \sim q \; \vee A_{n-(k-2)}$, so each time we get $\sim p$ and $\sim q$ which will be combined with already presented $\sim p$ and $\sim q$ and give $\sim p \; \vee \sim p = \sim p$ and $\sim q \vee \sim q = \sim q$ and at the end we get $ \sim p \; \vee \; \sim q \; \vee \; A_{something}$
If we keep doing this, at the end we get,
$$A_n =\; \sim p \; \vee \sim q \; \vee A_2$$
Since, $A_2 = \; \sim p \; \vee q$
Hence, $A_n = \; \sim p \; \vee \sim q \; \vee \; \sim p \; \vee q = \; \sim p \; \vee (\sim q \; \vee \; q) = True, n>2$