Ali who has $10$ candies eats at least one candy a day. In how many different ways can he eat all candies (according to distribution among days)?

$ 
\textbf{(A)}\ 64
\qquad\textbf{(B)}\ 126
\qquad\textbf{(C)}\ 243
\qquad\textbf{(D)}\ 512
\qquad\textbf{(E)}\ 1025
$
1. **Understanding the problem**: Ali has 10 candies and eats at least one candy each day. We need to find the number of different ways he can distribute the candies over several days.

2. **Formulating the problem**: Since Ali eats at least one candy each day, the problem can be seen as distributing 10 candies over a number of days such that each day has at least one candy.

3. **Using combinatorial reasoning**: We can think of this as a sequence of decisions. After eating each candy (except the last one), Ali can decide whether to continue eating candies on the same day or start eating on the next day. This decision is made 9 times (after each of the first 9 candies).

4. **Counting the choices**: Each of these 9 decisions has 2 options: either continue on the same day or start a new day. Therefore, the total number of ways to distribute the candies is given by:
   \[
   2^9
   \]

5. **Calculating the result**: 
   \[
   2^9 = 512
   \]

Thus, the number of different ways Ali can eat all the candies is 512.

The final answer is $\boxed{512}$