(x^2-5x+5)^(x^2-11x+30)=1⇒(x^2-5x+5)^((x-5)(x-6))=1. If x=5 or 6, the exponent becomes zero and (x^2-5x+5)^0=1.
If x^2-11x+30=1, we won't be able to factorise because we get x^2-11x+29=0; but if we move to x^2-11x+30=2, we get (x-4)(x-7)=0 so x is 4 or 7 and (x^2-5x+5)^2=1, so x^2-5x+5=1 or -1 and x^2-5x+4=0 or x^2-5x+6=0. This gives us (x-4)(x-1)=0 or (x-3)(x-2)=0. Therefore x=4 is a solution, but x=1, 2 or 3 are not (yet), because they produce the wrong exponent.
Up to now we have 3 solutions: x=4, 5, 6.
A pattern is emerging. x^2-11x+30=a where a is the exponent. The sum of the factors of this quadratic have to add up to 11 and their product has to be 30-a. We've had 5 and 6 when a=0; 4 and 7 when a=2. When a=6 we have factors 3 and 8, and (x^2-5x+5)^6=1; (x^2-5x+5)^3=1 or -1, so we know from earlier that this gives us potential factors 1, 2, 3, 4. Bingo! x=3 is another solution. Now we have four solutions: x=3, 4, 5, 6.
We have 9 and 2, and 10 and 1 as other factors summing to 11, and a=12 and 20. We already saw 1 and 2 as solutions, so now we have 6 solutions: 1, 2, 3, 4, 5, 6.