For a direct proof:
Assume $n$ is a positive integer (as it is false otherwise).
$\displaystyle \begin{align*}6^n + 4 & = 5\left(1+\sum_{k=0}^{n-1} 6^k\right) \\ & = 10\left(1 + \sum_{k=1}^{n-1} 3^k2^{k-1} \right)\end{align*}$
If $6^n$ has a unit digit of $6$ what is the unit digit of $6^n+4~?$
Look at this table.
@Stuck Man, I hope that you will tell us all what your instructor expected of you in an answer. Please.
Here are some comments I would ask you to consider.
You posted this question in a number theory forum. It is unthinkable that someone doing number theory would not have induction as a tool. As posed then this is basic algebra.
If I am correct about that, then I suspect the hint you were given, $6\cdot 6=~?$, is right on. At a very basic level, any integer ending in a six when multiplied by six results in an integer also ending in six. Now I admit that without induction there is no rigorous proof for that. But without induction available a rigorous proof is not called for.
There is a direct but very long proof.
$Step\ 1:\ Prove\ that\ 5m\ ends\ in\ 5\ if\ m\ is\ odd.$
$Step\ 2:\ Prove\ that\ 4^k\ is\ even.$
$Step\ 3:\ Prove\ that\ 5^p = (4 + 1)^p\ is\ odd.$
$Step\ 4:\ Prove\ that\ 5^q = 5 * 5^{(q - 1)}\ ends\ in\ 5.$
$Step\ 5:\ Prove\ that\ 6^r\ is\ even.$
$Step\ 6:\ Prove\ that\ 6^s = (5 + 1)^s\ ends\ in\ 1\ or\ 6.$
$Step\ 7:\ Prove\ that\ 6^n\ does\ not\ end\ in\ 1.$ This step is the only one that is even a bit tricky.
$Step\ 8:\ Prove\ that\ 6^n + 4\ ends\ in\ 0\ and\ so\ is\ divisible\ by\ 10.$ Trivial
I must admit that I have not worked out each step in detail, but I believe that you never need to use induction or rely on any theorem that itself is proved by induction. Step 6 does rely on contradiction. Can you use that?
AFTERTHOUGHT: I am not sure, but I think I see a way to avoid step 7. I'll think about it more as I walk the dog.
OK. I am back from letting the dog do canine things and think there is a way to prove this exercise without utlilizing reductio ad absurdum. I am not 100% sure. I was a history major, and my proofs are undoubtedly not up to modern standards of rigor.
Step 5 $\implies last\ digit\ of\ 6^n \in\ Set_A = \{0,\ 2,\ 4,\ 6,\ 8\}.$
Step 6 $\implies last\ digit\ of\ 6^n \in\ Set_B = \{1,\ 6\}.$
So Step 7 reduces to $last\ digit\ of\ 6^n \in Set_A \bigcap Set_B = \{6\}.$
Because I have never understood those mathematicians who deny the law of the excluded middle and consequently reject proofs by reductio, I don't know whether this is considered a direct argument or not, but it is not, at least not superficially, a proof by contradiction.
Hope this all helps.