Claim: a sequence $x(k)$ driven by simple state evolution with white "Gaussian" noise $n(k)$ below is a Markov sequence
\begin{align} x(0) &= 0 \\ x(k+1) &= x(k) + n(k), \quad k = 0, 1, 2, \dots, \end{align} where the white noise $n(k)$ has zero mean for all $k$, and covariance $\mathbb{E}[n(k)n^{\top}(j)] = C\delta_{kj},$ with Kronecker delta $\delta_{jk}$, and a constant $C$, where $C>0$, making the variance time-invariant resulting in a stationary white "Gaussian" noise. Additionally, $x(0)$ is also Gaussian.
How can I show that this is Markov?
The definition of Markov property I use is given below:
$$p[x(t)|x(\tau), \tau \leq t_1] = p[x(t)|x(t_1)] \quad \forall t>t_1$$
and I want to show: $p[x(k))|x(k-1), x(k-2), \dots, x(0)] = p[x(k)|x(k-1)]$.
Attempt:
\begin{align} p[x(k))|x(k-1), \dots, x(0)] &= p[x(k)|x(k-1), n(k-3), n(k-4), \dots, n(0)] \quad\text{$\dots(1)$}\\ &= \frac{p[x(k),x(k-1), n(k-3), n(k-4), \dots, n(0)]}{p[x(k-1), n(k-3), n(k-4), \dots, n(0)]}\quad\text{$\dots(2)$}\\ \end{align} For $(1)$, I make repetitive substitutions $x(k-2) = x(k-3) + n(k-3), x(k-3) = x(k-4) + n(k-4), \dots, x(k-k) = x(0) = 0.$
From $(2)$, I want to write:
$$p[x(k), x(k-1), n(k-3), \dots, n(0)] = p[x(k), x(k-1)]\cdot p[n(k-3)] \cdot \cdot \cdot p[n(0)]$$ on the numerator and $$p[x(k-1), n(k-3), n(k-4) , \dots, n(0)] = p[x(k-1)]\cdot p[n(k-3)] \cdot p[n(k-4)] \cdot \cdot \cdot p[n(0)]$$ on the denominator and cancel out, resulting in $$ p[x(k))|x(k-1), x(k-2), \dots, x(0)] = \frac{p[x(k), x(k-1)]}{p[x(k-1)]}\quad\text{$\dots(3)$}\\ $$
I am not sure if I am using a valid argument from $(2)$ to $(3)$ since the only fact I know about the white noise $n(k)$ is that its autocovariance is zero for two different times.
But does this fact also imply the independence of $n(k)$ and $n(j)$ for $k \neq j$ (which is what I am using)?
Edit after comments help
Generally, the noise in this process (discrete-time Wiener process) is drawn from iid Gaussian.
But I am not using Gausiann property anywhere and this raises a concern in my attempt. Is it necessary or can I drop the Gaussian assumption and still claim Markovness?
Otherwise, how should I use the "Gaussian" part in the proof for this process being Markov?