Here is my solution. Hope this helps! The answer is all numbers congruent to -1 mod 30. Checking, we get -1^-1 is congruent to -1 mod 30, so adding 1 yields divisibility by 30. Now we prove that this is the only solution. Since k^k+1 is even, k is odd. Now let us look at mod 3. k^k is congruent to 2 mod 3. When k is divisible by 3, k^k is also divisible by 3, so k is not divisible by 3. If k is congruent to 1 mod 3, then k^k is congruent to 1 mod 3, so k is not congruent to 1 mod 3. Therefore k is congruent to 2 mod 3. Checking, we see that when k is odd and congruent to 2 mod 3, k^k is also congruent to 2 mod 3. Now let us look at mod 5. k^k is congruent to 4 mod 5, so as mentioned before, k is not congruent to 0 or 1 mod 5. If k is congruent to 2 mod 5, then k^2, k^3 and k^4 are congruent to 4, 3 and 1 mod 5 respectively, hence it repeats with a period of 4. Since k is odd, k^k can only be congruent to 2 and 3 mod 5, which is a contradiction. We can do similar analysis with k being congruent to 3 mod 5. Hence, the only possibility left is k congruent to 4 mod 5, which works! Combining these three equations, we get k is congruent to -1 mod 30. The conclusion is proven.