The binomial theorem says
#(a+b)^n=sum_(k=0)^n(C_(n,k))(a^(n-k))(b^k)#, where
#C_(n,k)=(n!)/(k!(n-k)!)#
In this case
#a=m#
#b=(-a)#
and
#n=5#
So,
#(m+(-a))^5=sum_(k=0)^5(C_(n,k))(a^(n-k))(b^k)#
#=(C_(5,0))m^5(-a)^0+(C_(5,1))m^(5-1)(-a)^1+(C_(5,2))m^(5-2)(-a)^2#
#+(C_(5,3))m^(5-3)(-a)^3+(C_(5,4))m^(5-4)(-a)^4#
#+(C_(5,5))m^(5-5)(-a)^5#
#=(C_(5,0))m^5(-a)^0+(C_(5,1))m^4(-a)^1+(C_(5,2))m^3(-a)^2#
#+(C_(5,3))m^2(-a)^3+(C_(5,4))m^1(-a)^4#
#+(C_(5,5))m^0(-a)^5#
Since #x^0=1#, and #x^1=x# we can say
#=(C_(5,0))m^5(1)+(C_(5,1))m^4(-a)+(C_(5,2))m^3(-a)^2#
#+(C_(5,3))m^2(-a)^3+(C_(5,4))m(-a)^4#
#+(C_(5,5))(1)(-a)^5#
And since #(-x)(-x)=x^2#, and #(x^2)^2=x^4# we can also say
#=(C_(5,0))m^5+(C_(5,1))m^4(-a)+(C_(5,2))m^3(a)^2#
#+(C_(5,3))m^2(-a)^3+(C_(5,4))m(a)^4#
#+(C_(5,5))(-a)^5#
Also, on a similar note #(-x)(-x)(-x)=(-x)^3=-x^3#
and #(-x)^5=(-x)^3(-x)^2=(-x)^3(x^2)=-x^5#
So,
#=(C_(5,0))m^5-(C_(5,1))m^4a+(C_(5,2))m^3a^2#
#-(C_(5,3))m^2a^3+(C_(5,4))ma^4#
#-(C_(5,5))a^5#
Since, #C_(n,k)=(n!)/(k!(n-k)!)#, #0! =1# ,and #n! = prod_(k=1)^n k =1 times 2 times ... times n#
Then
#C_(5,0)=(5!)/(0!(5-0)!)=(5!)/((1)(5!))=1#
#C_(5,1)=(5!)/(1!(5-1)!)=(5!)/((1)(4!))=5#
#C_(5,2)=(5!)/(2!(5-2)!)=(5!)/((2)(3!))=(4 times 5)/2=20/2=10#
#C_(5,3)=(5!)/(3!(5-3)!)=(5!)/((3!)(2))=(4 times 5)/2=20/2=10#
#C_(5,4)=(5!)/(4!(5-4)!)=(5!)/((4!)(1))=5#
#C_(5,5)=(5!)/(5!(5-5)!)=(5!)/((5!)(0!))=(5!)/((5!)(1))=1#
So we plug in and we get
#(m-a)^5=m^5-5m^4a+10m^3a^2-10m^2a^3+5ma^4-a^5#
