Alphametic puzzle for year 2017

Question

In east Asia 2017 is the year of a rooster.

$$ABCD=\left(A \times CA^A + \frac{C}{D} \right) \times D$$

Answer 1

The solution is, of course,

A = 2, B = 0, C = 1, D = 7.

We can prove that this is a unique solution like so:

Looking at A,

A is the first digit of a 4-digit number, and so cannot be 0. A cannot be greater than 3 either, since 14⁴ is already 5 digits. A cannot be 3, since the only possible value for C keeping the result under 4 digits would be 1, but then there would be no solution for D. A cannot be 1, since then we would have (CD+C) equaling a 4-digit number, which is impossible. Therefore A must be 2.

Looking at C,

C is the first digit of a 2-digit number, and so cannot be 0. C cannot be greater than 3 either, since 2*42² is 3528, whose first digit is greater than 2. C cannot be 2, since A is 2. C cannot be 3, since 2*32² is 2048, and to keep the first digit 2, D is forced to be 1. However, subbing these values for A, C, and D in the right-hand side, the result is 2051, which doesn't fit the left-hand side. Therefore C must be 1.

Looking at D,

2*12² is 288. To get a 4-digit number starting with 2, D must be greater than or equal to 7. However, D cannot be 8 or 9 since the right-hand side would be 2305 or 2592 respectively, neither of which matches the left-hand side. D = 7 results in 2017, which fits and forces B to be 0.

Answer 2

With the question, as it is now.
This could be accepted

A,B,C = 0
D = 1

Answer 3

I don't know if there are several answers, but here is one possibility:

C is equal to 0, A and B are equal to any existing number, and D is equal to any number different of 0.