Given the binary operation $\star$ defined by $a\star b=a^b$ for all positive numbers $a$ and $b$. Then for all positive $a,b,c,n$, we have
$\text{(A) } a\star b=b\star a\quad\qquad\qquad\ \text{(B) } a\star (b\star c)=(a\star b) \star c\quad\\ \text{(C) } (a\star b^n)=(a \star n) \star b\quad \text{(D) } (a\star b)^n =a\star (bn)\quad\\ \text{(E) None of these}$
Let $a = 2, b = 3, c=n = 4$.  If all of them are false, the answer must be $E$.  If one does not fail, we will try to prove it.
For option $A$, we have $2^3 = 3^2$, which is clearly false.
For option $B$, we have $2^{81} = 8^{4}$, which is false.
For option $C$, we have $2^{81} = 16^3$, which is false.
For option $D$, we have $8^4 = 2^{12}$, which is true.
The LHS is $(a^b)^n$.  By the elementary definition of exponentiation, this is $a^b$ multiplied by itself $n$ times.  Since each $a^b$ is actually $a$ multiplied $b$ times, the expression $(a^b)^n$ is $a$ multiplied by itself $bn$ times.  
The RHS is $a^{bn}$.  This is $a$ multiplied by itself $bn$ times.
Thus, the LHS is always equal to the RHS, so $\fbox{D}$ is the only correct statement.