How do you simplify and write #(-5.3)^0# with positive exponents?

Remember the identity #a^m-:a^n=a^(m-n)#, for all #a#, where #m# and #n# are two natural numbers. For example

#(5.3)^3-:(5.3)^2=(5.3xx5.3xx5.3)/(5.3xx5.3)#

= #(cancel(5.3xx5.3)xx5.3)/(cancel(5.3xx5.3))#

= #5.3# and is nothing but #5.3^((2-1))=5.3^1=5.3#

Similarly #(-7.9)^5-:(-7.9)^3#

= #((-7.9)xx(-7.9)xx(-7.9)xx(-7.9)xx(-7.9))/((-7.9)xx(-7.9)xx(-7.9))#

= #((-7.9)xx(-7.9))=(-7.9)^2=(-7.9)^2=(-7.9)^((5-3))#

and hence #(-2.7)^3-:(-2.7)^3=(-2.7)^((3-3))=(-2.7)^0#

or #(-2.7)^3-:(-2.7)^3=((-2.7)xx(-2.7)xx(-2.7))/((-2.7)xx(-2.7)xx(-2.7))=1#

Hence for any number #a#,

if #m=n#, we get #a^m-:a^m=a^(m-m)#

or #a^m/a^m=a^(m-m)#

or #1=a^0#

Hence zero power of any number #a# is #1#

Hence #(-5.3)^0=1#.

Further demonstration of what is happening by example

Suppose we had #3^2/3^4#

Write as #(3xx3)/(3xx3xx3xx3)#

This is the same as: #3/3xx3/3xx1/3xx1/3" "=" "1xx1xx1/3^2#

Another way of writing #1/3^2" "# is #" "color(magenta)(3^(-2))#
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Lets look at this example again but in another way

write#" "3^2/3^4" as "3^2xx3^(-4)#

This can also be written as #" "3^(2-4)" which is the same as "color(magenta)(3^(-2))#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Answering your question")#

Consider #5^x# where #x# can by any whole number (integer)

Suppose we gave the value of 3 to #x# then we have#" "5^3#

Suppose we had #5^3/5^3# then by the method in the example I gave you we can write this as #" "5^(3-3)=5^0#

But #5^3/5^3=1#

So #1=5^3/5^3=5^(3-3)=5^0 = 1#

#color(magenta)("So a number raised to the power of 0 equals 1")#

#color(green)("In the question the minus is inside the bracket.")#
#color(green)("The index (power) of 0 is applied to everything inside")#
#color(green)("the bracket. So "color(magenta)((-5)^0=1))#
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Note that #-5^0->-(5^0)=-1#
Test this out on a calculator.