Hint: Observe $$(x+1)^2 - 1 = x(x+2),$$ or $$x+1 = \sqrt{1 + x(x+2)}.$$ Now recursively substitute: $$\begin{align*} x+1 &= \sqrt{1 + x\sqrt{1 + (x+1)(x+3)}} , \\ &= \sqrt{1 + x \sqrt{1 + (x+1)\sqrt{1 + (x+2)(x+4)}}}, \\ &= \sqrt{1 + x \sqrt{1 + (x+1)\sqrt{1 + (x+2)\sqrt{1 + (x+3)(x+5)}}}}, \ldots \end{align*}$$ Of course, you need to do something much more rigorous to complete the proof. I leave that to you.
Hint: Observe $$(x+1)^2 - 1 = x(x+2),$$ or $$x+1 = \sqrt{1 + x(x+2)}.$$ Now recursively substitute: $$\begin{align*} x+1 &= \sqrt{1 + x\sqrt{1 + (x+1)(x+3)}} , \\ &= \sqrt{1 + x \sqrt{1 + (x+1)\sqrt{1 + (x+2)(x+4)}}}, \\ &= \sqrt{1 + x \sqrt{1 + (x+1)\sqrt{1 + (x+2)\sqrt{1 + (x+3)(x+5)}}}}, \ldots \end{align*}$$ Of course, you need to do something much more rigorous to complete the proof. I leave that to you.