Every digit 0 thru 9 has a pattern of 4 digits when raised to an exponential power. Simply divide the power by 4 and the remainder shows you where you are at in the pattern. Here are the patterns for all 10 digits. Note that remainder 1 corresponds to the first digit in the pattern and remainder 2 corresponds to the second digit in the pattern. Remainder 3 corresponds to the third digit and remainder 0 (the power is divisible by 4) corresponds to the fourth digit in the pattern. Try a few yourself using a calculator and you will get the hang of it.
0,0,0,0
1,1,1,1
2,4,8,6
3,9,7,1
4,6,4,6
5,5,5,5
6,6,6,6
7,9,3,1
8,4,2,6
9,1,9,1
Being the curious kid that he is he asked why the two results weren't the same, and I couldn't give him an answer.
That's because the two results are the same, and he is implicitly using a slightly different and context-dependent notation to express his answer.
The arithmetic is correct, but $-4$ is not a decimal digit in the usual scheme of things.
A correct answer of $(-4).3$ was found, with an intended meaning of $-4 +0.3$. That notation is non-standard, and writing it as $-4.3$ gives the wrong answer when read as a standard decimal.
Although it's clear what an expression like $(-4).3$ should mean here, to represent that result in the standard system with digits 0-9, the minus sign can only apply to all digits in the number at once. The conversion to standard notation is $("-4").3 = -(3.7) = -3.7 $
Best Answer
$2^{4} = 16$. Multiply any even integer by $6$ and you don't change the last digit: $0 \times 6 = 0$, $2 \times 6 = 12$, $4 \times 6 = 24$ etc. The same is true if you multiply an even integer by anything whose last digit ends in $6$, in particular by $16$. Now $2006 = 2004 + 2$ where $2004 = 501 \times 4$, so $2^{2006} = (2^4)^{501} \times 2^2$ has the same last digit as $2^2$.