Proof by exhaustion: all positive integral powers of two end in 2, 4, 6 or 8 While learning about various forms of mathematical proofs, my teacher presented an example question suitable for proof by exhaustion:

Prove that all $2^n$ end in 2, 4, 6 or 8 ($n\in\mathbb{Z},n>0$)

I have made an attempt at proving this, but I cannot complete the proof without making assumptions that reduce the rigour of the answer.

All positive integral powers of two can be represented as one of the four cases ($k\in\mathbb{Z},k>0$, same for $y$):
  
  
*
  
*$2^{4k}=16^k=10y+6$
  
*$2^{4k+1}=2*16^k=10y+2$
  
*$2^{4k+2}=4*16^k=10y+4$
  
*$2^{4k+3}=8*16^k=10y+8$
  

The methods of proving the four cases above were similar; here is the last one:

$8*16^k=8*(10+6)^k$
Using binomial expansion,
$8*(10+6)^k=8*\sum\limits_{a=0}^k({k \choose a}10^k6^{k-a})$
All of the sum terms where $a\neq0$ end in zero, as they are a multiple of $10^k$, and therefore, a multiple of 10. The sum term where $a=0$ is $6^k$, because ${k\choose0}=10^0=1$. Therefore, the result of the summation ends in six.
Assuming that all positive integral powers of six end in six, and eight multiplied by any number ending in six ends in eight, all powers of two of the form $2^{4k+3}$ end in eight.

That conclusion doesn't seem very good because of the two assumptions I make. Can I assume them as true, or do I need to explicitly prove them? If I do need to prove them, how can I do that?

Hint $\ $ mod $10,\:$ the powers of $\:2\:$ repeat in a cycle of length $4,\:$ starting with $2,\:$ since
$$\rm  2^{K+4} = 2^K(1+15) = 2^K + 30\cdot2^{K-1}\equiv\: 2^K\ \ (mod\ 10)\quad for\quad K\ge 1$$
Now it suffices to prove by induction that if $\rm\:f:\mathbb N\to \mathbb N = \{1,2,3\ldots\}\:$ then
$$\rm  f(n+4)\: =\: f(n)\ \ \Rightarrow\ \  f(n)\in \{f(1),\:f(2),\:f(3),\:f(4)\}$$
Informally: once a cyclic recurrence begins to loop, all subsequent values remain in the loop.
Similarly, suppose there are integers $\rm\:a,b,\:$ such that $\rm\: f(n+2)\ =\: a\:f(n+1) + b\:f(n)\:$ for all $\rm\:n\ge 1.\:$ Show that $\rm\:f(n)\:$ is divisible by $\rm\:gcd(f(1),f(2))\:$ for $\rm\:n\ge 1$.
