I’m going to choose three subsets of $[n]$ as follows. First I choose any $i$ elements; call that set $A$. Then I choose $j$ of the remaining $n-i$ elements; call that set $B$. Finally, I choose $k$ of the $n-j$ elements in $[n]\setminus B$; call that set $C$. Now define a function $f_{A,B,C}:[n]\to[5]$ as follows:
$$f(k)=\begin{cases}
1,&\text{if }k\in A\setminus C\\
2,&\text{if }k\in A\cap C\\
3,&\text{if }k\in C\setminus A\\
4,&\text{if }k\in B\\
5,&\text{if }k\in[n]\setminus(A\cup B\cup C)
\end{cases}$$
There are
$$\sum_{k=0}^n\sum_{j=0}^n\sum_{i=0}^n\binom{n}i\binom{n-i}j\binom{n-j}k$$
ways to choose the sets $A,B$, and $C$, and each such choice uniquely determines a function $f_{A,B,C}:[n]\to[5]$. It’s clear that each $f:[n]\to[5]$ is $f_{A,B,C}$ for some choice of $A,B$, and $C$, and there are $5^n$ such functions, so
$$\sum_{k=0}^n\sum_{j=0}^n\sum_{i=0}^n\binom{n}i\binom{n-i}j\binom{n-j}k=5^n\;.$$
Here’s a bit of intuition. The expression on the lefthand side clearly suggests that we should first choose $i$ elements of $[n]$, the set that I called $A$, and then a set $B$ of $j$ elements disjoint from $A$. The third set, $C$, of $k$ elements apparently should be disjoint from $B$ but not necessarily from $A$. This clearly divides $[n]$ into five parts: the integers that are only in $A$, the integers that are in both $A$ and $C$, the integers that are only in $C$, the integers that are in $B$, and the integers that weren’t chosen for any of the three sets. we can tag those five sets with five colors, as in Parcly Taxel’s answer, or with the five elements of $[5]$, or with any handy $5$-element set to get the result.