S(n) = n^n + 1

Can the number S(n) = n^n+1 be prime? It is easy to show that a necessary condition is that n is of the form n=2^(2^k). The number S(n) therefore becomes a Fermat number F(m)=2^(2^m)+1 where m=k+2^k (Sloane's A006127).

Here is the status of such numbers. It seems extremely unlikely that any more of the numbers S(n) would be prime.

kmnstatus of S(n)
012prime, 2^2 + 1 = 5
134prime, 4^4 + 1 = 257
2616composite, 16^16 + 1 has factor 1071*2^8 + 1
311256composite, 256^256 + 1 has factor 39*2^13 + 1
42065536composite, but no factor known
537composite, S(n) has factor 1275438465*2^39 + 1
670status unknown
7135status unknown
8264status unknown
9521status unknown
101034status unknown
112059composite, S(n) has factor 591909*2^2063 + 1
124108status unknown
138205status unknown
1416398status unknown
1532783status unknown
1665552status unknown
17131089status unknown
18262162status unknown
19524307status unknown
201048596status unknown
212097173status unknown
224194326status unknown
238388631status unknown
2416777240status unknown
2533554457status unknown
2667108890status unknown

Links: Weisstein, Keller.