# Do the last digits of powers of a number $n$ follow the same cycle as the last digits of the number $n$'s last digit's powers?

In many places it is said that the last digits of the powers of the numbers from 1 to 9 have certain cycles. For example the last digits of powers of 2 repeat in a cycle of $4, 8, 6, 2$, and the last digits of powers of 9 repeat in a cycle of $1, 9$.

It seems like this works for bigger numbers as well. The last digits of any number's powers seem follow the cycle of the number's last digit's cycle. For example, the cycle of the last digits of powers of 7 is $9, 3, 1, 7$, and the cycle of the last digits of powers of 1097 are $9, 3, 1, 7$. I've been experimenting with my calculator and I haven't found a single counterexample, so my guess is that it's true for all numbers. That is, the last digits of powers of a number $n$ follow the same cycle as the last digits of the number $n$'s last digit's powers. Could someone show me a proof of this?

• The last digit of a power is completely determined by the last digit of the number. You can think of this as a consequence of the usual procedure for multiplying, but modular arithmetic shows this in a more satisfying and general way. Mar 26 '16 at 22:35

Your observation is correct. It's because, when you just look at last digits, you're working modulo 10, and addition and multiplication are well-defined modulo 10. Since, for example, $7\equiv 107\pmod{10}$, then for any natural $n$, we also have $7^n\equiv 107^n \pmod{10}$.
If you want technical details, write a number ending in the digit $b$ as $10k+b$. Then, if you raise that number to the power $n$, you can apply the binomial theorem:
$(10k+b)^n = (10k)^n + \binom{n}{1}\cdot(10k)^{n-1}b + \cdots + \binom{n}{n-1}(10k)b^{n-1} + b^n$
Since everything except for the last term is a multiple of 10, the last digit of this sum is simply the last digit of $b^n$.