jmc

Another way of writing the sequence $S$ is $\{7,7^7,7^{7^7},7^{7^{7^7}},\dots \}$. We wish to determine the $100^{\text{th}}$ term of this sequence modulo $5$.

Note that $s_{100}=7^{s_{99}}\equiv 2^{s_{99}}(\mathrm{mod}5)$. In order to determine the remainder of $2^{s_{99}}$ when divided by $5$, we look for a pattern in the powers of $2$ modulo $5$. Computing a few powers of $2$ yields $$\{2^0,2^1,2^2,2^3,2^4,\dots \}\equiv \{1,2,4,3,1,\dots \}(\mathrm{mod}5).$$So we have a cyclic pattern $1,2,4,3$ of length $4$ (this is called a period). Now we need to determine where $2^{s_{99}}$ falls in the cycle; to do that, we must determine the residue of $s_{99}(\mathrm{mod}4)$, since the cycle has length $4$.

Note that $$\begin{array}{rl}7& \equiv -1\equiv 3(\mathrm{mod}4),\\ 7^7& \equiv (-1{)}^7\equiv -1\equiv 3(\mathrm{mod}4),\\ 7^{7^7}& \equiv (-1{)}^{7^7}\equiv -1\equiv 3(\mathrm{mod}4),\\ & ⋮\end{array}$$Continuing in this manner, we always have $s_n\equiv 3(\mathrm{mod}4)$. Thus, $s_{100}=2^{s_{99}}\equiv 2^3\equiv \boxed{3}(\mathrm{mod}5)$.