Is the nth Catalan number smaller than n!?

Question

Can someone please provide both a formal proof? Also, given that the nth Catalan number is the number of binary trees with n nodes, is there a "physical" proof which involves reasoning about trees? Apologies if the 2nd part is vague.

Answer 1

Hint. The $n$th Catalan number counts the number of 2-symbol strings of length $2n$ with certain particular properties. Therefore it is obviously less than $2^{2n}=4^n$. This again is easy to compare to $n!$ for large enough $n$. Smaller $n$s can be checked manually.