The expression
$$ 1 + x + x^2 + x^3 + \dotsb = \sum_{k=0}^{\infty} x^k $$
is a geometric series. We say that a series converges if the corresponding sequence of partial sums converges. That is
$$ \sum_{k=0}^{\infty} x^k \text{ converges}\iff \lim_{n\to\infty} \sum_{k=0}^{n} x^k =: S_n < \infty.$$
But note that
$$ S_n = 1 + x + x^2 + x^3 + \dotsb + x^n
\implies x S_n = x + x^2 + x^3 + x^4 + \dotsb + x^{n+1}.$$
Subtracting and canceling terms as appropriate, this gives us
$$ (1-x)S_n = S_n - xS_n = 1 - x^{n+1}
\implies S_n = \frac{1-x^{n+1}}{1-x}.$$
Taking the limit, we get
$$ \lim_{n\to\infty} S_n
= \begin{cases}
\frac{1}{1-x} & \text{ if $|x|<1$ (since $x^{n+1} \to 0$),} \\
\infty & \text{ if $|x|>1$ (since $|x|^{n+1} \to \infty$), and} \\
\text{indeterminate} & \text{otherwise.}
\end{cases}
$$
Now, observe that if $|x|<1$, then
$$ f(x) = 1 + \sum_{k=2}^{\infty} x^k = \left(\sum_{k=0}^{\infty} x^k\right) - x = \frac{1}{1-x} - x.$$
For other values of $x$, the series won't converge, and the function will be undefined. Hence the domain of $f$ (presuming that the domain is real, and not complex) is the interval $(-1,1)$, and on this domain it is given by
$$ f(x) = \frac{1}{1-x} - x = \frac{1 - x(1-x)}{1-x} = \frac{1 - x + x^2}{1-x} = 1 + \frac{x^2}{1-x}.$$
Finally, suppose that $y$ is in the range of $f$. Then there is some $x\in (-1,1)$ such that $y=f(x)$. Solving, we get
$$ y = \frac{ 1-x + x^2}{1-x}
\implies (1-x)y = y - yx = 1-x+x^2
\implies 0 = x^2 + (y-1)x + (1-y).
$$
Thus
$$ x = \frac{1 - y \pm \sqrt{ (y-1)^2 - 4(1-y)}}{2}, $$
and so the equation $y = f(x)$ has a real solution $x$ as long as
$$ (y-1)^2 - 4(1-y)
= y^2 - 2y + 1 - 4 + 4y
= y^2 + 2y - 3
= (y+3)(y-1)
\ge 0. $$
This happens whenever $y\le-3$ or $y\ge 1$, i.e.
$$ y \in (-\infty,-3]\cup [1,\infty). $$
However, note that we also require $-1 < x < 1$. For such $x$, we have $1-x > 0$, and so
$$ y = 1 + \frac{x^2}{1-x} \ge 0.$$
Therefore the range of $f$ is contained in the set
$$ \Big[ (-\infty,-3]\cup [1,\infty) \Big] \cap [0,\infty)
= [1,\infty).$$
Slightly more carefully, note that $f$ is continuous on $(-1,1)$, that $f(0) = 0$, and that
$$\lim_{x\to 1^{-}} f(x) = \infty.$$
This is sufficient to show that $[1,\infty)$ is contained in the range of $f$, which finally allows us to conclude that the range of $f$ is
$[1,\infty).$
Something worth noting is that the domain of $f$ is the set $(-1,1)$, since $f$ is only defined when the series is convergent. However, as we observed above, on this domain, we have the identity
$$
f(x) = 1 + \sum_{k=2}^{\infty} x^k = 1 + \frac{x^2}{1-x}.
$$
The right-most expression defines a legitimate function on $\mathbb{R} \setminus \{1\}$. However, this is not the same function as $f$, since $f$ was defined by the series, and not by the rational expression on the right. We can, however, think of the function
$$ g(x) = 1 + \frac{x^2}{1-x} $$
as an extension of $f$ to a larger domain. Indeed, in the context of complex analysis (a branch of mathematics that deals with functions that have domains in the complex plane), it can be shown that this extension is (in some sense) the "right" extension.
