2 typo.; added 20 characters in body

This is not a full answer but I'm throwing some observations out. Note that the even stronger inequality also has a clear combinatorial interpretation. Namely, if one denotes by inj(A,B) the set of injections from A to B, then the inequality says that

|inj(T,inj(W,V))| > |inj(W,inj(T,V))|.

To see this, multiply both sides of the stronger inequality by t!w!. Of course, here t = |T|, w = |W| and v = |V|.

In this form, it is easy to believe the inequality. Consider a function $f : T \times W \to V$, thought of as a $t \times w$ matrix with entries in V. It defines an element of the left hand side if there is no repeated entry in any row, and there are no repeated columnsentire rows. Vice versa for the The right hand side counts the same thing with columns. We have assumed that the columns are longer than the rows. Then if the entries of the matrix are picked uniformly at random, it should fail to define an element of the right hand side with greater probability, since most functions that fail should do so because of repeated entries in a row/column rather than an entire repeated row/column, and it is more likely to be a repeated entry in a column than in a row.

However, I don't see any nice way of producing a proof out of the above heuristic. Bijective proofs seem to get really complicated...

1

This is not a full answer but I'm throwing some observations out. Note that the even stronger inequality also has a clear combinatorial interpretation. Namely, if one denotes by inj(A,B) the set of injections from A to B, then the inequality says that

|inj(T,inj(W,V))| > |inj(W,inj(T,V))|.

To see this, multiply both sides of the stronger inequality by t!w!. Of course, here t = |T|, w = |W| and v = |V|.

In this form, it is easy to believe the inequality. Consider a function $f : T \times W \to V$, thought of as a $t \times w$ matrix with entries in V. It defines an element of the left hand side if there is no repeated entry in any row, and there are no repeated columns. Vice versa for the right hand side. We have assumed that the columns are longer than the rows. Then if the entries of the matrix are picked uniformly at random, it should fail to define an element of the right hand side with greater probability, since most functions that fail should do so because of repeated entries in a row/column rather than an entire repeated row/column, and it is more likely to be a repeated entry in a column than in a row.

However, I don't see any nice way of producing a proof out of the above heuristic. Bijective proofs seem to get really complicated...