Why is $2^{n}-1=\sum\limits_{k=0}^{n-1}2^k$? Recently I discovered that $2^{n}-1=\sum\limits_{k=0}^{n-1} 2^k$. As I don't have a math background, please tell me what this is called and a proof of why this is the case.

As Hayden has pointed out, the correct expression is $2^n-1=\sum_{k=0}^{n-1} 2^k.$
So to see this, let’s first name the sum $S:=1+2^1+\dots+2^{n-2}+2^{n-1}.$
The ‘trick’ is to multiply the sum by $1$ in a clever way such that we can cancel some terms. Since $2=2-1$, we can say this:
$$
S=1+2^1+\dots+2^{n-2}+2^{n-1}\\
\implies\\
1\times S = \left(2-1\right)\times \left(1+2^1+\dots+2^{n-2}+2^{n-1}\right)\\
S=2\times\left(1+2^1+\dots+2^{n-2}+2^{n-1}\right)\\
-1\times\left(1+2^1+\dots+2^{n-2}+2^{n-1}\right)\\
= \left(2^1+2^2+\dots+2^{n-1}+2^{n}\right)\\
-\left(1+2^1+\dots+2^{n-2}+2^{n-1}\right)\\
= 2^n-1.
$$
Hope this helps!
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