PARITY $\notin$ $AC^0$ with bounded fanout: easy proof? $AC^0$ is the class of constant-depth polynomial-size circuits with NOT gates and  unbounded fan-in AND and OR gates, where inputs and gates also have unbounded fanout.

Now consider a new class, call it $AC^0_{bf}$ which is like $AC^0$ but for which inputs and gates have fanout at most $O(1)$.  This class is clearly in $AC^0$.  In fact, it is strictly contained in $AC^0$, as noted here.  Therefore, PARITY is obviously not in $AC^0_{bf}$.

Is there a proof of PARITY $\notin AC^0_{bf}$ which does not also go through for $AC^0$? In other words, is there a proof which does not use powerful techniques like the switching lemma or the Razborov/Smolensky method?
I might miss something, but isn't $AC^0_{bf}$ the same as a Formula? Since every input bit can have an effect on at most a bounded number of gates, we can simply suppose that every gate has only one output (after possibly duplicating a few things) and we can push down not gates as well. We know that the formula size of parity is n^2 (see Troy J. Lee, "The formula size of PARITY", 2007) and since on every level of our circuit we can only have O(n) gates, this shows that parity is not in $AC^0_{bf}$.