Solve by using modulo. Or something else An infinite sequence of positive integers $a_1, a_2,\ldots$ has the properties that for $k\geq2$, the $k^\text{th}$ element is equal to $k$ plus the product of the first $k-1$ elements of the sequence. Suppose $a_1 =1$. what is the smallest prime number that does not divide $a_{2010}$?
I have a feeling you have to use mod? But how do you find the twenty tenth number of the sequence?

Modular arithmetic is congruent under multiplication, so we can define sequences $a_k^{(2)}$, $a_k^{(3)}$, $a_k^{(5)}$, etc. to represent the sequences modulo 2, 3, and 5 respectively.
Thanks to congruency we still have the recurrence:
$$a_k^{(p)} \equiv \prod_{i=1}^{k-1}a_i^{(p)} + k \mod p$$
Now it remains to look at patterns in sequences $a^{(p)}_k$.
Modulo 2:
Note:
$$a_3^{(2)} \equiv 6 \equiv 0 \mod 2$$
So for any $k>3$, the product term is zero mod 2. Hence we simplify for large $k$ to:
$$a_k^{(2)} \equiv k \mod 2$$
So 2 divides $a_{2010}$ since $2010 \equiv 0 \mod 2$.
Mod 3:
Same approach:
$$a_3^{(3)} \equiv 6 \equiv 0 \mod 3$$
Hence, with the same reasoning,
$$a_{2010}^{(3)} \equiv 2010 \equiv 0 \mod 3$$
So 3 is not the factor we are looking for.
Mod 5:
First notice:
$$a_1^{(5)} a_2^{(5)} \equiv 1 \times 3 \equiv 3 \mod 5$$
Let's list out the next few terms:
$$
\begin{split}
a_3^{(5)} &\equiv 3 + 3 \equiv 1 &\mod 5 \newline
a_4^{(5)} &\equiv 3 \cdot 1 + 4 \equiv 2 &\mod 5 \newline
a_5^{(5)} &\equiv 3 \cdot 1 \cdot 2 + 5 \equiv 1 &\mod 5 \newline
a_6^{(5)} &\equiv 3 \cdot 1 \cdot 2 \cdot 1 + 6 \equiv 2 &\mod 5 \newline
a_7^{(5)} &\equiv 3 \cdot 1 \cdot 2 \cdot 1 \cdot 2 + 7 \equiv 4 &\mod 5
\end{split}
$$
Now let's look at $a_8^{(5)}$.
$$a_8^{(5)} \equiv 3 \cdot 1 \cdot 2 \cdot 1 \cdot 2 \cdot 4 + 8 \equiv 3 + 3 \mod 5$$
Note how the product term has reset itself to 3 mod 5! This means we have entered into a cycle, since the recurrence into $a_9^{(5)}$ will be like the recurrence into $a_4^{(5)}$, and so on. Indeed, $a_{k+5}^{(5)} = a_{k}^{(5)}$ for all sufficiently large $k$. Notice how the remainder never reaches 0, so 5 is our final answer.
