I recommend to use the following inequality,Originally Posted by The Great One
a^x-a^{x-1} <= a^x - 1 <= a^x
Thus,
(a^x-a^{x-1})^{1/x} <= (a^x - 1)^{1/x} <= a
Right? I did not check this yet and about to sleep.
I know where you get that problem from. It was from the Putnam. (I have a Calculus book which has Putnam problems in it and I remember I did this one).
The one they have is actually stronger version.
Let a!=0 and a>0.
---
I do not know why this call this "beauty limit". If the limit was something like pi^a, then that would be a beauty limit.