Here's another approach.

Consider f(n) = 11 x 14^n + 1, modulo 15.

Then f(n) = 11 x (-1)^n + 1 = 12 or 5 (mod 15).

Hence f(n) = 0 (mod 3) or 0 (mod 5), is greater than 5 for all n, and is always composite.

(f(n) = 12 (mod 15) <=> f(n) = 15t + 12, for some integer t <=> f(n) = 3(5t + 4) <=> f(n) = 0 (mod 3), etc.)

Or you could use induction.

f(0) = 12 is divisible by 3,

f(1) = 155 is divisible by 5.

Then f(n+2) - f(n) = 11 × 14^n(14^2 - 1) = 11 × 3 × 5 × 13 × 14^n is divisible by both 3 and 5, and the result follows by induction.

(f(n) divisible by 3 implies f(n+2) divisible by 3; f(n) divisible by 5 implies f(n+2) divisible by 5.)

The digital roots always go in cycles of 6.

That is there are six possible digital roots (sometimes, one or more of them may be the same)Since the digital root of 14 is 5,

we get 6 values of digital roots for

5^6n, 5^6n+1, 5^6n+2,....5^6n+5

They are 1, 5, 7, 8, 4, 2.

For even powers, we see that the digital roots are 1, 7, 4.

This would be true for 5, 14, 23, 32, 41 etc, i.e. all numbers having digital root of 5.

From the work of Mathsy, it is seen that we needed to check only for even powers (as odd powers are divisible by 5).

Since the even powers are divisible by 3, it can be concluded that

14^n + 11 would never be a prime number!

I don't think there are any loopholes in the proof.

For the even values, it seems as if the digital root of 14^n is always 1, 4 or 7, which means that 14^n+11 will always be divisible by 3 when n is even, but I don't see how I can prove that. If that can be proved, then the answer to the original question can be proved to be no.

Note: The digital root is when all the digits of a number are added together until you only have one digit.

e.g. 77-->7+7=14-->1+4=5.