My response was as follows:
We may test our base case, that is when $k=0$. This computes to $x^{0}=\frac{1-x^{1}}{1-x}$ which equates to $1=1$. Whence $k=0$ holds. Therefore, such a statement $P_{k} :=x\neq1$ and k ${\in}\> \textbf{Z}_{\geq0}$,$\>\sum_{j=0}^{k}x^{j}=\frac{1-x^{k+1}}{1-x}$ is assumed to be true.
Subsequently, our inductive hypothesis is formulated in evaluating the statement $P_{k+1}$. When prompted with such a dilemma, what may be useful is to write out the individual elements of the sequence to be added on our left hand side. For example, $$\>\sum_{j=0}^{k+1}x^{j}=\frac{1-x^{k+2}}{1-x}$$ may be written as $$x^{1}+x^{2}+x^{3}+\dots+x^{k-1}+x^{k}+x^{k+1}=\frac{1-x^{k+2}}{1-x}$$ This may be further made intuitive when we evaluate both sides as such and multiply $(1-x)$ by our left hand side expression. $$(x^{0}+x^{1}+x^{2}+\dots+x^{k-1}+x^{k}+x^{k+1})\cdot(1-x)=1-(x\cdot x \cdot x \cdot x^{k-1})$$ Then $$(x^{0}+x^{1}+x^{2}...+x^{k-1}+x^{k}+x^{k+1})-(x^{1}+x^{2}+x^{3}+\dots+x^{k}+x^{k+1}+x^{k+2})=1-(x\cdot x \cdot x \cdot x^{k-1})$$ Finally this rationalizes to $x^{0}-x^{k+2}=1-x^{k+2}$ which completes are induction step.
The statement $P_{k+1}$ holds as $1-x^{k+2}=1-x^{k+2}$ is true. By the principle of mathematical induction our original equality remains. ////
Is my solution adequate? Would further proof be needed to show that this holds for $x \neq 1$?
Your proof by induction is correct. However, it is a little convoluted, in my opinion. Here is how I would do it by induction.
The verification of the base case $k = 0$ is fine as it stands.
The induction hypothesis is formulated as follows: if $k \in \mathbb{Z}_{> 0}$ and $x \neq 1$, then $$\sum_{j=0}^k x^j = \frac{1-x^{k+1}}{1-x}.$$
Now, we want to check that for $x\neq 1$ and $k \in \mathbb{Z}_{>0}$, we have $$\sum_{j=0}^{k+1} x^j = \frac{1-x^{k+2}}{1-x}.$$ Working with the left hand side, we get \begin{align} \sum_{j=0}^{k+1} x^j &= \left( \sum_{j=0}^k x^j \right) + x^{k+1}\\ &= \frac{1-x^{k+1}}{1-x} + x^{k+1}\\ &= \frac{1-x^{k+1} + x^{k+1} - x^{k+2}}{1-x}\\ &= \frac{1-x^{k+2}}{1-x}. \end{align} Hence, by the principle of mathematical induction, we have proved our claim.
You can also prove the claim directly, without using induction. Let $$S = \sum_{j=0}^k x^j.$$ Then, $$xS = \sum_{j=1}^{k+1} x^j.$$ Hence, $$S - xS = x^0 - x^{k+1} \implies S = \frac{1-x^{k+1}}{1-x}.$$