Well Bertie I have given it a shot 
$$\\\mbox{Prove }\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{.....}}}}=x+1\\
\\LHS\\
f(x)=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{.....}}}}\\
f(x+1)=\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{.....}}}\\\\
f(x)=\sqrt{1+x*f(x+1)}\\\\
f^2(x)=1+x[f(x+1)]$$
$$\\RHS\\
g(x)=x+1\\
g(x+1)=x+1+1\\
g(x+1)=1+g(x)\\
g^2(x)=(x+1)^2\\
g^2(x)=x^2+2x+1\\
g^2(x)=1+x(x+2)\\
g^2(x)=1+x[g(x+1)]\\$$
$$\\f^2(x)=1+x[f(x+1)]\qquad and\qquad g^2(x)=1+x[g(x+1)]\\\\
$It follows that$\\\\
f^2(x)=g^2(x)\\\\
Now f(x)\ge 0 \qquad $because it is the answer to a square root$\\\\
g(x)=x+1 $ which \;must\; be\; $\ge 0 $ Therefore x is greater than or equal to -1$\\\\$$
$$\mbox{The statement cannot be true if x is less than -1 because the}\\ \mbox{RHS would be neg and the Left hand side can never be negative}$$
$$\\$so for all real values of x greater than or equal to -1$\\
\\f(x)=g(x) \\\\
\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{.....}}}}=x+1\qquad Q.E.D.\\$$
As an aside:
$$\\\mbox{A look at two individual cases }\\
If\;\; x=0 \;then\\
LHS=\sqrt{1+0*...}=\sqrt1=1=RHS\\\\
If\;\; x=-1 \;then\\
LHS=\sqrt{1+-1*\sqrt{1+(-1+1)\sqrt...}}=
\sqrt{1+-1*\sqrt{1+0}}=\sqrt{1+-1*1}=\sqrt{0}=0=RHS\\\\$$
I am really not completely sure of of this proof.
Is it really true for ALL real values greater of equal to -1?
Is it completely correct Bertie or does it need refining?
+130511 
