Step 1: Choose the units digit route instead of induction.
Rather than proving the pattern by induction, we can settle this just by tracking what happens to the last digit when we multiply repeatedly by 6.
Step 2: State the rule for units digits.
When two numbers are multiplied, the units digit of the product depends only on the units digits of the two numbers being multiplied, nothing else about the numbers matters. So if a number already ends in 6, multiplying it by 6 again only depends on what $6 \times 6$ ends in.
\[ 6 \times 6 = 36 \]
This ends in 6. So once a number ending in 6 gets multiplied by 6, the result still ends in 6.
Step 3: Set the starting point and repeat the rule.
For $n = 1$, $6^1 = 6$, which ends in 6.
For $n = 2$, we multiply the previous result (ending in 6) by 6 again, and by the rule above the units digit stays 6, giving $6^2 = 36$.
For $n = 3$, we multiply 36 (ending in 6) by 6, so the units digit is 6 again, giving $6^3 = 216$.
Every time we go from $6^k$ to $6^{k+1}$, we are multiplying a number ending in 6 by 6, and the rule from Step 2 tells us the units digit never changes from 6. Since this repeats forever no matter how large $n$ gets, the units digit is locked at 6 for every natural number $n$.
Step 4: Final answer.
$6^n$ always ends with the digit 6 for every natural number $n$, which is option (B).
\[ \boxed{6} \]