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I have encountered this question in a coding competition and want the formula for $f(n)$. Is there any way to do this?

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    You should probably specify the first couple of terms... e.g. $f(1)$ and $f(2)$. – 2017-01-19
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    $1+f(n)=(1+f(n-1))(1+f(n-2))$ – 2017-01-19
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    Welcome to the site! Please make sure your question is self-contained and that all relevant information is in the question itself, and not just the title. Also, let us know what you have tried so far and where to you get stuck or lost. – 2017-01-19
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    Could you provide a link to the competition? Is it still on-going? – 2017-01-20

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We have from @Daniel's comment,

$$\ln (1+f(n))=\ln((1+f(n-1))(1+f(n-2))$$

$$\ln (1+f(n))=\ln (1+f(n-1))+\ln (1+f(n-2))$$

Let $A(n)=\ln (1+f(n))$ then we have,

$$A(n)=A(n-1)+A(n-2)$$

This has characteristic equation,

$$r^2-r-1=0$$

Whose solution is the golden ratio $\phi$ and $1-\phi$.

Hence,

$$A(n)=c_1(\phi)^n+c_2(1-\phi)^n$$

$$1+f(n)=\exp (c_1(\phi)^n+c_2(1-\phi)^n)$$

$$f(n)=\exp (c_1(\phi)^n+c_2(1-\phi)^n)-1$$

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Supposing the first two terms are $0$ and $1$ (or $1$ and $1$) this would be OEIS entry A063896:

$f(n) = 2^{F(n)} - 1$ where $F(n)$ is the $n$th Fibonacci Number.

To prove that this relation continues to hold, you may wish to try mathematical induction.