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}