Induction involving $x$ is no different from other types of induction. Just treat $x$ normally and we can still prove it for the general case.
Let $$P_n:=\sum_{i=0}^nx^i=\frac{1-x^{n+1}}{1-x}$$
Then for the base case i.e. $n=0$ we have
$$\sum_{i=0}^0x^i=\frac{1-x^{1+0}}{1-x}$$
$$1=\frac{1-x}{1-x}$$
$$1=1$$
He have esablished the RHS=LHS and so $P_0$ is true.
Now assume that the statement is true for n.
$$\sum_{i=0}^nx^i=\frac{1-x^{n+1}}{1-x}$$
We must show that it is also true for $n+1$.
For the RHS we have
$$RHS=\frac{1-x^{(n+1)+1}}{1-x}=\frac{1-x^{n+2}}{1-x}$$
And for the LHS we have
$$LHS=\sum_{i=0}^{n+1}x^i=\sum_{i=0}^{n}x^i+x^{n+1}=\frac{1-x^{n+1}}{1-x}+x^{n+1}=\frac{1-x^{n+1}}{1-x}+\frac{x^{n+1}(1-x)}{(1-x)}$$
$$=\frac{1-x^{n+1}+x^{n+1}(1-x)}{1-x}=\frac{1-x^{n+1}+x^{n+1}-x^{n+2}}{1-x}=\frac{1-x^{n+2}}{1-x}$$
Since LHS=RHS $P_{n+1}$ is true. The rest follows.