It's from the sum of a (finite) geometric series. But you can derive it from first principles.

$$S_n(x) = 1 + x + x^2 + \dotsb + x^n$$

$$xS_n(x) = x + x^2 + x^3 + \dotsb + x^{n+1}$$

Subtracting the second from the first (and noting the telescoping nature, which I'm making explicit here),

$$(1-x)S_n(x) = 1 - x + x - x^2 + x^2 + \dotsb - x^n + x^n - x^{n+1} = 1- x^{n+1}.$$

Rearranging,

$$S_n(x) = \frac{1-x^{n+1}}{1-x}.$$

Observe that $$ \frac{1}{x-1}(x^{k+1}-x^{k})=x^k\quad (x\neq 1) $$ whence $$ \sum_{k=0}^n x^k=\sum_{k=0}^n\frac{1}{x-1}(x^{k+1}-x^{k})=\frac{1}{x-1}(x^{n+1}-1) =\frac{1-x^{n+1}}{1-x};\quad (x\neq 1) $$ since the sum telescopes.

$S_n(x)=1+x+x^2+x^3+ . . .x^n=1+x+x^2+x^3+ . . .x^n +x^{n+1}-x^{n+1}=1-x^{n+1} + x(1 +x+x^2+x^3 . . .+x^n)=1-x^{n+1} +x S_(n)$

⇒ $(1-x)S_n(x)=1-x^{n+1}$

⇒ $S_n(x)=\frac{1-x^{n+1}}{1-x}$

For completeness let me add one, also very usual, argumentation. Partial sum can be derived from formula: $$a^{n}-b^{n}=(a-b)(a^{n-1} + ba^{n-2}+ \cdots + b^{n-1}) $$ Taking $b=1$ we obtain $$a^{n}-1=(a-1)(a^{n-1} + a^{n-2}+ \cdots + 1) \Rightarrow a^{n-1} + a^{n-2}+ \cdots + 1 = \frac{a^{n}-1}{a-1}$$

Deepak's excellent answer is the standard argument. I will give (essentially) the same argument here, but a slightly different presentation. What I like about this argument, in comparison to Deepak's, is that it eliminate the ellipses and makes the computations a little more precise. There is a cost—I think that some of the intuition is lost, since we don't see the term-by-term cancelation—but I think that this is a price which can be paid without too much difficulty.

Given any real (or complex) number $x$ and any natural number $n$, let $S_n(x)$ denote the $n$-th partial sum of the series $\sum_{j=0}^{\infty} x^j$. That is, $$ S_n(x) = \sum_{j=0}^{n} x^j. $$ Observe that \begin{align} xS_{n}(x) - S_{n}(x) &= x\sum_{j=0}^{n} x^j - \sum_{j=0}^{n} x^j \\ &= \sum_{j=0}^{n} x^{j+1} - \sum_{j=0}^{n} x^j &&\text{(distribution over finite sums)} \\ &= \sum_{k=1}^{n+1} x^{k} - \sum_{j=0}^{n} x^j &&\text{(CoV: let $k=j+1$)} \\ &= \left[ \sum_{k=1}^{n} x^k + x^{n+1}\right] - \left[1 + \sum_{j=1}^{n} x^j \right] && \text{(pull out a couple of terms)} \\ &= x^{n+1} + \color{red}{\sum_{k=1}^{n} x^k} - \color{red}{\sum_{j=1}^{n} x^j} - 1 && \text{(the red terms cancel)} \\ &= x^{n+1} - 1. \end{align} Supressing the intermediate steps, this reduces to \begin{align} x S_n(x) - S_n(x) = x^{n+1} - 1 &\implies (x-1)S_n(x) = x^{n+1} - 1 \\ &\implies S_n(x) = \frac{x^{n+1}-1}{x-1} = \frac{1-x^{n+1}}{1-x}, \end{align} which is the claimed identity.

Another alternative to Deepak's appeal to telescoping sums is the following. Again, we start with \begin{align} (x-1)S_n(x) &= xS_n(x) - S_n(x) \\ &= \left[ x + x^2 + x^3 + \dotsb + x^n + x^{n+1} \right] - \left[ 1 + x + x^2 + x^3 + \dotsb + x^n \right]. \end{align}

If we write this subtraction in the style that is taught in American elementary schools, it looks something like \begin{array}{r} &&& \color{red}{x} &+& \color{blue}{x^2} &+& \color{green}{x^3} &+& \dotsb &+& \color{orange}{x^{n}} &+& x^{n+1} \\ -{\quad} & 1 &+& \color{red}{x} &+& \color{blue}{x^2} &+& \color{green}{x^3} &+& \dotsb &+& \color{orange}{x^{n}} \\\hline & -1 &+& \color{red}{0} &+& \color{blue}{0} &+& \color{green}{0} &+& \dotsb &+& \color{orange}{0} &+& x^{n+1}. \end{array} Lining up "like terms" (that is, aligning terms with the same exponent) makes it a little easier to see where the cancelations are happening. The rest of the argument is identical.