Consider $\{Y(t), t \geq 0\}$ and $Y(t) = \exp\{cB(t) - c^{2}t/2 \}$, where $c$ is a constant and $\{B(t), t \geq 0\}$ is a standard Brownian motion process.
Why is it the case that $$e^{-c^{2}t/2}E[e^{cB(t)} | B(u), 0 \leq u \leq s] = e^{-c^{2}t/2} E[e^{cB(t)}|B(s)]$$
This kind of thing is usually argued by considering $B(s) + B(t) - B(s)$ and then using the independent increment property of Brownian motion. But what justifies this property in this case?
Use independent increments again. If $s\le t$ and $\mathcal F_s$ denotes $\sigma(B(u), 0\le u \le s)$, then $$ \begin{align} E[e^{cB(t)} | \mathcal F_s] &= E[e^{c(B(t)-B(s))}e^{cB(s)} | \mathcal F_s] \\&\stackrel{(1)}= E[e^{c(B(t)-B(s))}]E[e^{cB(s)} | \mathcal F_s] \\&\stackrel{(2)}= E[e^{c(B(t)-B(s))}]e^{cB(s)}\\&\stackrel{(3)}= E[e^{c(B(t)-B(s))}]E[e^{cB(s)}|B(s)]\\&\stackrel{(4)}= E[e^{c(B(t)-B(s))}e^{cB(s)} | B(s)] \\&= E[e^{cB(t)} | B(s)] \end{align} $$ Equalities (1) and (4) follow from independent increments; equality (2) follows because $B(s)$ is $\mathcal F_s$-measurable.
This result is a special case of:
If $X$ is $\mathcal H$-measurable and $Y$ is independent of $\mathcal H$, then $\mathbb{E}(g(X,Y)|\mathcal{H})=\mathbb{E}(g(X,Y)|X)$.
More details in this post. The fact a process with independent increments is Markov also follows from the more general result.
