Is the Equation x^{a^b} = (x^{a^{b-1}})^a True for All Natural Numbers?

[notnottrue]

Hi,
If all x,a,b and c are all natural numbers, is this true?
$x^{a^b} = (x^{a^{b-1}})^a$
Proof
if $c = a^{b-1}$
$ca = (a^{b-1})a = a^b$
and $(x^c)^a = x^{ca} = x^{a^b}$

Could I please have some feedback on this,
Thanks

 

[pwsnafu]

There is no universal agreement on whether zero is a natural number. You are going to need write "positive integer" or deal with the case when one or a number of the variables are zero separately.

 

[Mentallic]

Hi,
If all x,a,b and c are all natural numbers, is this true?
$x^{a^b} = (x^{a^{b-1}})^a$

Yes, and the proof doesn't require any substitutions either. Simply following your exponent laws,

$$\left(x^{a^{b-1}}\right)^a$$
$$=x^{a^{b-1}a}$$
$$=x^{a^{b-1+1}}$$
$$=x^{a^b}$$