One knows that #a^2=axxa#, #a^3=axxaxxa# and so on i.e. #a^m=axxaxxaxx...xxa#, where #a# is multipled by itself #m# times.
Now howwe divide say #a^7# by #a^4#.
Well #a^7=axxaxxaxxaxxaxxaxxa# and #a^4=axxaxxaxxa#
Hence #a^7-:a^4=a^7/a^4=(axxaxxaxxaxxaxxaxxa)/(axxaxxaxxa)#
= #(color(red)(cancelaxxcancelaxxcancelaxxcancela)xxaxxaxxa)/(color(red)(cancelaxxcancelaxxcancelaxxcancela))#
= #axxaxxa#
= #a^3#
Hence we can write #a^7-:a^4=a^(7-4)=a^3#
Observe that for all #m# and #n#, we can have #a^m/a^n=a^(m-n)#
Can we write #a^0=a^(m-m)=a^m/a^m=1#?
Answer is yes we can and hence #a^0=1#, where #a# isany number. Hence any number raised to power zero is one.
Additional information #-# Note that #a^4-:a^7=a^4/a^7=(axxaxxaxxa)/(axxaxxaxxaxxaxxaxxa)#
= #(color(red)(cancelaxxcancelaxxcancelaxxcancela))/(color(red)(cancelaxxcancelaxxcancelaxxcancela)xxaxxaxxa)#
= #1/(axxaxxa)#
= #1/a^3#
Can we write #a^(-3)=a^(4-7)=1/a^3#?
Answer is yeswe can and #a^(-m)=1/a^m#
