Question 122260
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Use the rule {{{b^m^n=b^mn}}}, and repeat it three times to get 


{{{x^0^2^4^5=x^(0*2*4*5)}}}


But {{{0*2*4*5=0}}} so your expression becomes


{{{x^0}}}


But {{{a^0=1}}} for all real a.


So, {{{x^0^2^4^5=1}}}


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