Always the Same Last Digit

If some integer $Y$ is divided by 10, the remainder will always be the last digit of $Y$ . So the question becomes; what integer $D$ , no matter the power it is raised to, has a constant last digit? To answer this we look at the table

We see that the last digit of powers of $1$ 's, $5$ 's and $6$ 's are all constant. Now, we have $3$ integers $D$ that satisfy the property, but we must note that 0 (though not included in the table) may be a possible solution and obviously it is, as $0^N=0$ for all positive integer values of $N$ .

Therefore, there are $\boxed{4}$ integers which satisfy the property.

$D^N$ is constant iff $D^N \equiv D^M$ (mod 10), for every ordered pair of positive integers (N, M).

Now $D^2 \equiv D$ (mod 10).

By trial and error we find that D can be 0,1,5 or 6, and thus there are $\boxed{4}$ integers which satisfy this property.

All number can be written in the form $(10k+b)^n$ and by binomial expansion, the unique element that isn't multiplied by $10$ is the last digit. Therefore when you divide the number by 10 element per element, the unique element of the binomial expansion that is not necessary divisible by 10 is the last digit, thus if the last digit is always equal independent of the exponent, the remainder will be equal, i.e we need that the last digit of power will be equal. This is in 0,1,5,6

0,1, 5, and 6 are the only solutions.