Reduce 4-SAT to 5-SAT
Here is a translation of the answer how to prove 4-SAT CNF is NP-complete to the current situation.



Suppose an instance of 4-SAT over variables $x_1,x_2,\cdots, x_m$ is given as a boolean formula $$f=c_1\land c_2\land \cdots\land c_m,$$ where $c_i$ is a disjunction that has exactly 4 literals for all $i$.

Introduce a new variable $s$. Suppose $c_i=w\lor x \lor y \lor z$ for some literal $w,x,y,z$. Let 
$$\begin{aligned}
c_{i+}&amp;=w\lor x \lor y \lor z\lor s\\
c_{i-}&amp;=w\lor x \lor y \lor z\lor \neg s.
\end{aligned}$$


If $c_i$ can be satisfied by an assignment, then $c_{i+}\land c_{i-}$ is satisfied by the same assignment plus $s=0$ (or $s=1$). 
If $c_{i+}\land c_{i-}$ can be satisfied by an assignment, the same assignment without considering $s$ must satisfy $c_i$ since one of $s$ and $\neg s$ must be false. 


Construct an instance of 5-SAT over variables $x_1,x_2,\cdots, x_m, s$, $$g=c_{1+}\land c_{1-}\land c_{2+}\land c_{2-}\land\cdots\land c_{m+}\land c_{m-}.$$

Because of the relation between $c_i$ and $c_{i+}\land c_{i-}$,


If $f$ is satisfied by an assignment, then $g$ is satisfied by the same assignment plus $s=0$. 
Conversely, if $g$ is satisfied by an assignment, then $f$ is satisfied by the same assignment without considering $s$. 


The above means the transformation from $f$ to $g$ is a reduction from 4-SAT to 5-SAT. It runs in polynomial time.



For simplicity of explanation, the above restricts $k$-SAT to CNF formulas with exactly $k$ literals without duplicates. Thanks to a similar simple padding technique, it is does not matter to the construction of reduction above whether we allow at most $k$ literals with or without duplicates when we define $k$-SAT.