User blog comment:SuperJedi224/Question/@comment-1605058-20141124134651

I think it's clear that \(\beta(n)<\Gamma_0\), so \(\beta(\omega)\leq\Gamma_0\). Now, if \(\alpha,\gamma\leq\beta(n)\), then \(\varphi(\alpha,\gamma)\leq\varphi(\beta(n),\beta(n))\leq\varphi(\beta(n)+1,0)=\beta(n+1)\), with last inequality holding because, first fixed point of \(\varphi(\alpha,x)\) is surely at least \(\alpha\). So \(\beta(\omega)\) is closed under \(\varphi\) function, so \(\beta(\omega)\geq\Gamma_0\). So \(\beta(\omega)=\Gamma_0\).