Prove for any integer $a$ show that $a$ and $a^{4n+1}$ have the same last digit 
For any integer $a$, show that $a$ and $a^{4n+1}$ have the same last digit

I know that if $a^{4n+1} \equiv a\pmod{10}$ then $10|a^{4n+1}-a$, so $2|a^{4n+1}-a$ and $5|a^{4n+1}-a$, but I'm not sure where to go from here.
 A: You're right: you want to prove that $a^{4n+1}\equiv a\pmod{10}$, which is equivalent to proving that
$$
a^{4n+1}\equiv a\pmod{2}
\qquad\text{and}\qquad
a^{4n+1}\equiv a\pmod{5}
$$
This should be easier, because you can apply
$$
a^4\equiv 1\pmod{5}
$$
for every $a$ not divisible by $5$.
A: This relies on the concept captured by the Carmichael function, $\lambda(10)=4$, the least universal exponent. 
All numbers will enter a cycle of values under exponentiation$\bmod m$. $\lambda(m)$ captures how long that cycle can be and requires that the actual cycle length will divide that number.
Since $\lambda(10)=4$, values repeat on a cycle of $4$, and since the highest prime power that divides $10$ is a prime, you can be sure that the cycle has always been entered by $a^1$. 
For the last two digits, using $\bmod 100$ which has $\lambda(100)=20$,  we can equivalently say that $a^2\equiv a^{20k+2}\bmod 100$, with $a^2$ required because $100$ is divisible by a prime square (but not by a prime cube), so we need to take two exponentiation steps to saturate the prime powers.
A: Here's a more elementary argument.  Note that mod $10$ we have
$$a^{4n+1}-a = a(a^{4n}-1) = a(a^{2n}-1)(a^{2n}+1) \equiv a(a^{2n}-1)(a^{2n}-9) =a(a^n-1)(a^n+1)(a^n-3)(a^n+3) \pmod{10}.$$
If $\gcd(a,10)=1$ or $10$, then $a^n \equiv 0, \pm1 \mbox{ or }\pm3 \pmod{10}$, making the above product congruent to $0$ mod $10$.  
If $\gcd(a,10) = 5$, then $a$ is odd and each of the other factors in the product is even, so again the product is congruent to $0$ mod $10$.
If $\gcd(a,10) = 2$, then $a^n \equiv 2, 4, 8, \mbox{ or } 6 \pmod{10}$  Each of these is $\pm 1$ or $\pm 3$ from $5$, so $5$ is a factor of the product and again the product is congruent to $0$ mod $10$.
A: The most elementary method is simply to observe patterns of the last digits ( there are only 10 possible digits to check so it is quite fast)  by multiplying last digit by $a,a^2,a^3,a^4 $, for greater powers - the pattern is repeating ..
$0 \rightarrow 0,0,0,0,.. $
$1 \rightarrow 1,1,1,1,.. $
$2 \rightarrow 4,8,6,2,.. $
$3 \rightarrow 9,7,1,3,.. $
$4 \rightarrow 6,4,6,4,.. $
$5 \rightarrow 5,5,5,5,.. $
$6 \rightarrow 6,6,6,6,.. $
$7 \rightarrow 9,3,1,7,.. $
$8 \rightarrow 4,2,6,8,.. $
$9 \rightarrow 1,9,1,9,.. $   
From this you can devise even the more general proposition - if $a$ and $b$ have the same last digit then $ab^{4m}$ or $ba^{4m}$ have the same last digit like $a$ and $b$.
A: If $\gcd(a,10) = 1$ then $a^{\phi(10)} = a^4 \equiv 1\mod 10$.  So $a^{4n+1} = (a^4)^n*a\equiv 1^na \equiv a \mod 10$.
If $\gcd(a,10) \ne 1$ then $a = 0, 5$ or $2^kb; \gcd(b,10) =1$.
$0^k = 0\equiv 0 \mod 10$ (duh) and $5^k \equiv 5 \mod 10$ (do I really need to show that?)
$6^2 = 36\equiv 6 \mod 10$ so by induction $6^k \equiv 6 \mod 10$.
So if $a = 2^k*b$ then $$a^{4n+1} = a *(a^{4n})=a*((2^4)^{kn}b^4)\equiv 16^{kn}a\equiv 6^{kn}a$$$$ \equiv 6a\equiv a(5+1) \equiv 5a + a \equiv 10*\frac a2 + a \equiv a\mod 10$$.
