A basic power problem. 
What is unit digit of $14^{15^{16^{17}}}$? (no brackets given)

This question is simple and it is based on cyclicity, but the confusion arise when I taken this approach:
We know that $5$ raised to any natural number will give $5$ in the unit place. Which will make the number odd.
And the cyclicity of $4$ is $2$, which is $(4,6)$ so by this concept I got the answer $4$.
But my friend first resolved $14^{15}$ which will give $4$ as unit digit. Again, that unit digit $4$ raised to $16$ will give $6$ as unit place and again $6$ raised to $17$ will return $6$ as unit digit.So by this way answer is $6$.
How do we resolve $a^{b^c}$? Do we find $a^b$ and then raise it to $c$? Or first we resolve $b^c$ and raise $a$ to it?
P.S :
This is my first post on stack exchange. Please let me know if I had done anything wrong with description.
Thanks in advance.


So basically to cut the long story short how do we resolve a^b^c ? Do we find a^b and the the answer of that is raised to c? Or first we resolve b^c and the answer is raised to a?

Most people who have an opinion uses $b^c$ first, then raise $a$ to the result. Calculating $a^b$, then raise that to the power of $c$ is the same as $a^{bc}$, so it makes sense to let a^b^c mean the other one.
I personally consider a^b^c to be ambiguous and ill-defined.
