Lower bound of sorting by comparison is O(n * log n)

Found a really nice explanation on why there cannot be a sorting algorithm that uses comparisons and can have a lower bound then O(n * log n).

An Ω(nlogn) lower bound can be shown by observing that any sorting algorithm must behave differently during execution on each of the distinct n! permutations of n keys. The outcome of each pairwise comparison governs the run-time behavior of any comparison-based sorting algorithm. We can think of the set of all possible executions of such an algorithm as a tree with n! leaves. The minimum height tree corresponds to the fastest possible algorithm, and it happens that lg(n!) = Θ(n log n).