Recurrence relations: prove that if $a_i = a_j$, then $i=j$ 
let $$a_n = \frac{n^2+10}{10^4}a_{n-1}, \space a_1=99$$
  Prove (or disprove) that $a_i = a_j \implies i = j$

My proof:
The above sequence is non-monotonic: $\frac {a_n}{a_{n-1}} > 1$ when $n^2 > 9900$. Therefore the least element will be $a_{99}$. Since a difinite minimum element exists, it is possible that if $a_i = a_j$, then $i=j$ exists such that $i<99, j>99$. 
This proof is not only weak but also wrong: my book says that this property is true. Please help me in proving so.
 A: Hint: If $a_i = a_j$ where WLOG $i < j$, then $$1 = \dfrac{a_j}{a_i} = \prod_{n = i+1}^{j}\dfrac{a_n}{a_{n-1}} = \prod_{n = i+1}^{j}\dfrac{n^2+10}{10^4} = \dfrac{1}{10^{4(j-i)}}\prod_{n = i+1}^{j}(n^2+10)$$ This means that we need $$\prod_{n = i+1}^{j}(n^2+10) = 10^{4(j-i)}.$$ Now, for what values of $i,j$ can the above product be a power of $10$, let alone $10^{4(j-i)}$?
A: Well, this is the same as proving that $f:\mathbb{Z^{+}}\rightarrow\mathbb{Z^{+}}$ is injective $(f(a)=f(b) \implies a=b)$
while $f(1)=99$ and 
$$f(x)=\frac{x^2+10}{10^4}\cdot f(x-1)$$
$$f(x)=\frac{x^2+10}{10^4}\cdot\frac{(x-1)^2+10}{10^4}\cdot \dots \cdot\frac{(x-k)^2+10}{10^4} \cdot f(x-k-1) \tag{1}$$
for all positive integers $k \le x-2$
Assume for the sake of contradiction that there exists positive integers $a,b$ such that $f(a)=f(b) \ne0$ while $a \ne b$, and assume WLOG that $a>b$, so that we can choose $k$ such that $b=a-k-1$, we get:
$$f(a)=\frac{a^2+10}{10^4}\cdot\frac{(a-1)^2+10}{10^4}\cdot \dots \cdot\frac{(b+1)^2+10}{10^4} \cdot f(b)$$
$$\iff(a^2+10)((a-1)^2+10)\dots((b+1)^2+10)=1$$
which is, of course, impossible.
