I want to show that $\text{NP} \neq \text{SPACE}(n)$ and tried it like this:
Let $L$ be in $\text{SPACE}(n)$ so there is a deterministic $k$-tape TM which decides $L$ in polynomial time. Let's consider a binary alphabet. There are $2^{kn}$ possible strings on the working tapes so for an input of length $n$ the TM can be in $2^{kn} \cdot k \cdot n \cdot |Q|$ different configurations where $Q$ is the finite state set.
Since the TM is deterministic and decides $L$ no configuration can be visited twice (otherwise we would be in an infinite loop). So $2^{kn} \cdot k \cdot n \cdot |Q|$ is also an upper bound for the runtime of the TM which is exponential and therefore not in $NP$.
Is this argumentation enough? I found this answer on the internet and now I'm confused because it looks so much more complexive but also comprehensibly...