The sequence is recursively defined by
[tex]\begin{cases}a_1 = 1 \\ a_2 = 2 \\ a_n = a_{n-1} a_{n-2} & \text{for } n \ge 3\end{cases}[/tex]
By this definition,
[tex]a_{n-1} = a_{n-2} a_{n-3} \implies a_n = a_{n-2}^2 a_{n-3}[/tex]
[tex]a_{n-2} = a_{n-3} a_{n-4} \implies a_n = (a_{n-3} a_{n-4})^2 a_{n-3} = a_{n-3}^3 a_{n-4}^2[/tex]
[tex]a_{n-3} = a_{n-4} a_{n-5} \implies a_n = (a_{n-4} a_{n-5})^3 a_{n-4}^2 = a_{n-4}^5 a_{n-5}^3[/tex]
[tex]a_{n-4} = a_{n-5} a_{n-6} \implies a_n = (a_{n-5} a_{n-6})^5 a_{n-5}^3 = a_{n-5}^8 a_{n-6}^5[/tex]
and so on.
Recall the Fibonacci sequence, {1, 1, 2, 3, 5, 8, 13, 21, …}, where the next term in the sequence is the sum of the previous two terms. If [tex]F_n[/tex] is the n-th Fibonacci number, then continuing the pattern above we would arrive at
[tex]a_n = {a_2}^{F_{n-1}} {a_1}^{F_{n-2}} = 2^{F_{n-1}}[/tex]
Notice that the sequence of positive powers of 2 leaves a periodic sequence of residues mod 10 :
[tex]2 \equiv 2 \pmod{10}[/tex]
[tex]2^2 \equiv 4 \equiv 4 \pmod{10}[/tex]
[tex]2^3 \equiv 8 \equiv 8 \pmod{10}[/tex]
[tex]2^4 \equiv 16 \equiv 6 \pmod{10}[/tex]
[tex]2^5 \equiv 2 \times 2^4 \equiv 2 \times 6 \equiv 2 \pmod{10}[/tex]
[tex]2^6 \equiv 2^2 \times 2^4 \equiv 4 \times 6 \equiv 4 \pmod{10}[/tex]
and so on; the period of this sequence of residues is 4.
The period of [tex]F_n[/tex] taken mod 4 is 6 :
[tex]\{1, 1, 2, 3, 5, 8, \ldots\} \equiv \{1, 1, 2, 3, 1, 0, \ldots\} \pmod 4[/tex]
(This follows from the "properties" section in the link in comment. In this case, π(4) = 3/2 × 4 = 6.)
It follows that
[tex]34 \equiv 4 \pmod 6 \implies F_{34} \equiv 3 \pmod 4 \implies 2^{F_{34}} \equiv 2^3 \equiv 8 \pmod{10}[/tex]
which means the last digit of [tex]a_{35}[/tex] is 8.
