The sum of the first $n$ powers of $\frac1s$ is given by the formula for a geometric progression:
$$\sum_{k=0}^{n}ar^k=\frac{a(1-r^n)}{1-r} $$
For the progression $1+ \frac{1}{s} + \frac{1}{s^2} + \frac{1}{s^3} + ... + \frac{1}{s^n}$, we have $a=1 $ and $ r=\frac1s$, and the above formula simplifies to $$\frac{s(1-\frac1{s^n})}{s-1}$$
Thus, $$1+ \frac{1}{s} + \frac{1}{s^2} + \frac{1}{s^3} + ... + \frac{1}{s^n} - 1 \\ =\frac{1}{s} + \frac{1}{s^2} + \frac{1}{s^3} + ... + \frac{1}{s^n} = \frac{s(1-\frac1{s^n})}{s-1}-1=\frac{1-\frac{1}{s^{n-1}}}{s-1} $$
But we can also simplify $\frac{1}{s} + \frac{1}{s^2} + \frac{1}{s^3} + ... + \frac{1}{s^n}$ by directly using the geometric progression formula with $a=r=\frac1s$, which gives $$\frac{1-\frac{1}{s^{n}}}{s-1} $$
Which is different than $\frac{1-\frac{1}{s^{n-1}}}{s-1} $. Which formula is correct, and where is my mistake?
Your formula $$\sum_{k=0}^{n}ar^k=\frac{a(1-r^n)}{1-r}$$
Should have been $$\sum_{k=0}^{n}ar^k=\frac{a(1-r^{n+1})}{1-r}$$