Find an exact expression for the n-th derivative of the function $f(x) = e^{e^{x}}$.
I took a few derivates of the expression and found out that a pattern emerges such that the n-th derivative will be expressed as $$S(1, n)* e^xe^{e^{x}} + S(2,n)*e^{2x}e^{e^{x}} + S(3,n) * e^{3x}e^{e^{x}}+ S(4,n) * e^{4x}e^{e^{x}} +.... +S(n,n) e^{nx}e^{e^{x}}$$
where S(n,k) is a Stirling number of the second kind.
My question is how would I find the exact expression? Do I have to prove this someway?
Once you suspect this pattern, namely that $$\frac{d^n}{dx^n}\left[e^{e^x}\right] = e^{e^x} \sum_{k=1}^n S(n,k) e^{kx},$$ then a proof by induction would be a natural choice for demonstrating its validity. (I have used the more standard ordering $S(n,k)$ rather than your $S(k,n)$.)
You may also be interested in the Touchard polynomials.