Question 860335
You want to prove that
{{{2^1+2^2+2^3+"..."+2^n=2^(n+1)-2}}}
or, using fancier symbols, that
{{{sum(2^p,p=1,p=n)=2^(n+1)-2}}} . 
You need to prove that
a) it is true for {{{n=1}}} , and
b) if it is true for {{{n=k}}} , then it will be true for {{{n=k+1}}} .
 
For {{{n=1}}} the "sum" has only one term:
{{{sum(2^p,p=1,p=n)=sum(2^p,p=1,p=1)=2^1=2}}}
and it is indeed equal to
{{{2^(n+1)-2=2^(1+1)-2=2^2-2=4-2=2}}}
 
If the formula {{{2^1+2^2+2^3+"..."+2^n=2^(n+1)-2}}} is true for {{{n=k}}} ,
{{{2^1+2^2+2^3+"..."+2^k=2^(k+1)-2}}} .
Adding one more term, we find
{{{2^1+2^2+2^3+"..."+2^k+2^(k+1)=2^(k+1)+2^(k+1)-2=2*2^(k+1)-2=2^((k+1)+1)-2}}} ,
which shows that the formula is true for {{{n=k+1}}} .