If $x=1$, then its powers end with $1$ How to "show", that if $x=1$, then its powers will end with $1$?
Or, if $x=5$, how do we show it's powers end with $5$?
 A: I am presuming you are talking about the last digit of a number in base $10$ being $1$ and $5$ implying the powers of that number will also end with $1$ and $5$.
I will give you a hint:
Give two numbers whose last digits are $1$, what does their product end with?
A: If you know modular arithmetic: $\rm\; e\equiv 1,5\;\Rightarrow\; e^2\equiv e \;\Rightarrow\; e^k\equiv e\pmod{10}\;$ for all $\rm\; k>0$
If not, use induction: suppose $\rm\;\; e^2 = e + 10j\;$ and  $\rm\; (e + 10 m)^k = e + 10n$
then $\rm\; (e + 10 m)^{k+1} = (e + 10n) (e + 10m) = e + 10(j+em+en+10mn)$
REMARK Note how the use of modular arithmetic has allowed us to rephrase the problem in a completely arithmetical form that makes the induction step utterly trivial - namely that $\rm 1^k \equiv 1$ or, more generally that $\rm\; e^2 \equiv e\;\Rightarrow e^k \equiv e\;$ for $\rm\: k>0$. This is but one very simple illustration of how one may often reduce problems about integers to simpler problems about integers modulo $\rm\:m\:$. Because the integers modulo $\rm\:m\:$ satisfy many of  the same arithmetical laws as the integers (associative, commutative, distributive, etc) - i.e. they form a ring - we can reuse our strong intuition about integers to help discover proofs in this simpler structure integers modulo $\rm\:m\:$. And, if we're lucky, we can sometimes lift these simple solutions from the integers modulo $\rm\:m\:$ up to the integers. If you study abstract algebra you'll see many more examples of problems solved by passing to such simpler structures (technically they are known as quotient objects - structures modulo a congruence, i.e. a structure-compatible equivalence relation). Said in an informal way, such reduction to quotient objects is a method to divide and conquer mathematically.
A: Use mathematical induction:


*

*Show that $x^1$ ends in 1 (or 5).

*Show that if $x^n$ ends in 1 (or 5), then $x^{n+1}$ also ends in 1 (or 5).  (Think about multiplying a number than ends in 1 by 1 (or ends in 5 by 5).)

A: If you know that addition and multiplication are well-defined modulo $n$, then just work (mod 10).
Let $x_0 \in$ {1,5} be the ones digit of $x$.  Then for any $k$
$x^k \equiv x_0^k \equiv x_0$ (mod 10).
