It's easy to see that $ a_{n} > 0,(\forall)n\geq 1.$ $ a_{n + 1} = \frac {a_{n}\cdot a_{n - 1}}{\sqrt {a_{n}^2 + a_{n - 1}^2 + 1}}\Longleftrightarrow a_{n + 1}^2\cdot a_{n}^2 + a_{n + 1}^2\cdot a_{n - 1}^2 + a_{n + 1}^2 = a_{n}^2\cdot a_{n - 1}^2.$ Let $ \{x_{n}\}_{n\geq 1}$ such that $ x_{n} = a_{n}^2.$ So $ x_{n} > 0, (\forall)n\in\mathbb{N}^*,$ and $ x_{1} = 1;x_{2} = 25.$ $ a_{n + 1}^2\cdot a_{n}^2 + a_{n + 1}^2\cdot a_{n - 1}^2 + a_{n + 1}^2 = a_{n}^2\cdot a_{n - 1}^2\Longleftrightarrow$ $ x_{n + 1}\cdot x_{n} + x_{n + 1}\cdot x_{n - 1} + x_{n + 1} = x_{n}\cdot x_{n - 1}$ $ |\cdot\frac {1}{x_{n - 1}\cdot x_{n}\cdot x_{n + 1}}\Longleftrightarrow$ $ \Longleftrightarrow \frac {1}{x_{n - 1}} + \frac {1}{x_{n}} + \frac {1}{x_{n - 1}}\cdot \frac {1}{x_{n}} = \frac {1}{x_{n + 1}}.$ Consider the sequence $ \{y_{n}\}_{n\geq 1}$ defined by: $ y_{n} = \frac {1}{x_{n}} = \frac {1}{a_{n}^2}$ and $ y_{1} = 1;y_{2} = \frac {1}{25}.$ Now: $ \frac {1}{x_{n - 1}} + \frac {1}{x_{n}} + \frac {1}{x_{n - 1}}\cdot \frac {1}{x_{n}} = \frac {1}{x_{n + 1}}\Longleftrightarrow$ $ y_{n - 1} + y_{n} + y_{n - 1}\cdot y_{n} = y_{n + 1}$ $ | + 1\Longleftrightarrow$ $ (y_{n - 1} + 1)(y_{n} + 1) = y_{n + 1} + 1.$ Let's make the substitution: $ z_{n} = y_{n} + 1,(\forall)n\in\mathbb{N}^*.$ We have to determine the general term of sequence $ \{z_{n}\}_{n\geq 1}$ defined by: $ z_{n + 1} = z_{n}\cdot z_{n - 1},(\forall)n\geq 2$ and $ z_{1} = 2;z_{2} = \frac {26}{25}.$ Of course $ z_{n} = z_{1}^{p_{n}}\cdot z_{2}^{q_{n}}.$ Let us define other two sequences: $ p_{n} = e_{z_{1}}(z_{n})$ and $ q_{n} = e_{z_{2}}(z_{n}),(\forall)n\in\mathbb{N}^*.$ We get the recurrences: $ p_{n + 1} = p_{n} + p_{n - 1},(\forall)n\geq 4$ with $ p_{3} = 1;p_{4} = 1$ and $ q_{n + 1} = q_{n} + q_{n - 1},(\forall)n\geq 4$ with $ q_{3} = 1;q_{4} = 2.$ Then we obtain that: $ p_{n} = \frac {3\sqrt {5} - 5}{10}\cdot (\frac {1 + \sqrt {5}}{2})^n - \frac {5 + 3\sqrt {5}}{10}\cdot (\frac {1 - \sqrt {5}}{2})^n$ and $ q_{n} = \frac {5 - \sqrt {5}}{10}\cdot (\frac {1 + \sqrt {5}}{2})^n + \frac {5 + \sqrt {5}}{10}\cdot (\frac {1 - \sqrt {5}}{2})^n,(\forall)n\geq 3.$ So $ a_{n} = \sqrt {\frac {1}{z_{n} - 1}} = \sqrt {\frac {1}{z_{1}^{p_{n}}\cdot z_{2}^{q_{n}} - 1}},(\forall)n\geq 3.$

We let $F_1=1,F_2=0,F_n=F_{n-1}+F_{n-2}(n\ge3)$ $F_n$ is the well-known Fibonacci Sequence,with $F_n=\frac{1}{\sqrt{5}}[(\frac{1+\sqrt{5}}{2})^{n-2}-(\frac{1-\sqrt{5}}{2})^{n-2}]$ Then from the formula we get $1+\frac{1}{a_n^2}=(1+\frac{1}{a_{n-1}^2})(1+\frac{1}{a_{n-2}^2})$ So by induction we can get $1+\frac{1}{a_n^2}=(1+\frac{1}{a_1^2})^{F_n}(1+\frac{1}{a_2^2})^{F_{n+1}}=2^{F_{n+2}}13^{F_{n+1}}5^{-2F_{n+1}}$ So $a_n=(2^{F_{n+2}}13^{F_{n+1}}5^{-2F_{n+1}}-1)^{-\frac{1}{2}}$