Let $P(n)$ be the assertion. Then $P(n+1)-P(n)$ gives $f(n+1)^2-f(n)f(n+2)=1$. Also $P(1)$ gives $f(1)=1$. Let $Q(n)$ be the assertion that $f(n+1)^2-f(n)f(n+2)=1$ then $Q(2)$ and $Q(3)$ gives $f(2)^2-f(3)=1$ and $f(3)^2-4f(2)=1$. $f(2)=a$ gives $f(3)=a^2-1$ then $(a^2-1)^2-4a=1\Rightarrow a(a-2)(a^2+2a+2)=0\Rightarrow f(2)=2$. I claim for all $n$ that $f(n)=n$. If it is true for $n$ and $n+1$ then by $Q(n)$ it is true for $n+2$ too. It is true for $1$ and $2$ so by induction we are done.

TheMaskedMagician wrote: Find all functions $f:\mathbb{N}\rightarrow(0,\infty)$ such that $f(4)=4$ and \[\frac{1}{f(1)f(2)}+\frac{1}{f(2)f(3)}+\cdots+\frac{1}{f(n)f(n+1)}=\frac{f(n)}{f(n+1)},~\forall n\in\mathbb{N},\]where $\mathbb{N}=\{1,2,\dots\}$ is the set of positive integers.

Solution: Replace $n$ by $n+1$, then substract, and we get $f(n)f(n+2)+1={{f}^{2}}(n+1)$ We will prove by induction that $f(n)=n$. Suppose that $f(k)=k$, and $f(k+1)=k+1$, we will prove that $f(k+2)=k+2$. This is indeed true, because: $f(k)f(k+2)+1={{f}^{2}}(k+1)\Leftrightarrow kf(k+2)+1={{(k+1)}^{2}}$