Does the product rule for exponents apply to a power of a power? Does the product rule for exponents apply to a power of a power?
For example: If I have 3^(4^5), will that be equal to 3^(4^2)*3^(4^3)?
 A: First, note that exponents are evaluated right-to-left. What I mean by this is that
$$
{a^b}^c
$$
by convention means $a^{(b^c)}$. It is important to have such a convention, since
$$
a^{(b^c)} \neq (a^b)^c
$$
in general. For example,
$$
2^{(3^4)} \neq (2^3)^4  \, .
$$
Because of this, your question as it stands is a little ambiguous. When you write 'a^b^c', do you mean $${(a^b)}^c$$ or $$a^{(b^c)} \, ?$$ If you mean the former, then the product rule for exponents does hold:
$$
(a^b)^c \times (a^b)^d = (a^b)^{c+d} \, .
$$
To explain why, try setting $k=a^b$. Then we have
$$
k^c \times k^d = k^{c+d} \, ,
$$
which is the familiar product formula. However, if by a^b^c you meant
$$
a^{(b^c)} \, ,
$$
then the product formula does not apply in the way you suggested. Instead,
$$
a^{b^c} \times a^{b^d} = a^{b^c+b^d} \, .
$$
Try working out why this is the case, and ask me if you have any questions.
A: It depends on how you write it.
$$3^{(4^5)}\ne 3^{(4^2)}.3^{(4^3)}$$
but
$$(3^4)^5=(3^4)^2.(3^4)^3$$
A: It applies in these terms: $$3^{4^2}\times 3^{4^3}=3^{4^2+4^3}\stackrel{\text{assuming you like this}}=3^{4^2(4^{3-2}+1)}=3^{5\cdot4^2}$$
And I don't think you want claim that $3^{4^5}=3^{80}$. I sure don't.
A: A short answer is no. Using the product rule that works we have
$$3^{4^2}3^{4^3} = 3^{4^2+4^3}\neq 3^{4^5} $$
