Let us take a particular example that is large enough to illustrate the general situation. Concrete experience should precede the abstract.
Let $n=8$. We want to show that $2^0+2^1+2^2+\cdots +2^8=2^9-1$. We could add up on a calculator, and verify that the result holds for $n=8$. However, we would not learn much during the process.
We will instead look at the sum written backwards, so at
A kangaroo is $2^9$ feet from her beloved $B$. She takes a giant leap of $2^8$ feet. Now she is $2^8$ feet from $B$. She takes a leap of $2^7$ feet. Now she is $2^7$ feet from $B$. She takes a leap of $2^6$ feet. And so on. After a while she is $2^1$ feet from $B$, and takes a leap of $2^0$ feet, leaving her $2^0$ feet from $B$.
The total distance she has covered is $2^8+2^7+2^6+\cdots+2^0$. It leaves her $2^0$ feet from $B$, and therefore
Since $2^0=1$, we obtain by subtraction that $2^8+2^7+\cdots +2^0=2^9-1$.
We can write out the same reasoning without the kangaroo. Note that
$2^0+2^0=2^1$, $2^1+2^1=2^2$, $2^2+2^2=2^3$, and so on until $2^8+2^8=2^9$.
Subtract the front $2^0$ from the left side, and $2^0$, which is $1$, from the right side, and we get our result.