If $\varphi \left(x\right)$ is a polynomial function and  ${\varphi }^{\mathrm{\prime }}\left(x\right)>\varphi \left(x\right)\mathrm{\forall }x\ge 1$   and  $\varphi \left(1\right)=0$, then

Given ${\varphi }^{\mathrm{\prime }}\left(x\right)-\varphi \left(x\right)>0\mathrm{\forall }x\ge 1$

or $e^{-x}\left\{{\varphi }^{\mathrm{\prime }}\right(x)-\varphi (x\left)\right\}>0\mathrm{\forall }x\ge 1$

or $\frac{d}{dx}{e}^{-x}\varphi \left(x\right)>0\mathrm{\forall }x\ge 1$

Therefore, $e^{-x}\varphi \left(x\right)$ is an increasing function  $\mathrm{\forall }x\ge 1$.

$\text{since }\varphi \left(x\right)$ is a polynomial,

$e^{-x}\varphi \left(x\right)>e^{-1}\varphi \left(1\right) \text{or }{e}^{-x}\varphi \left(x\right)>0$

$[\because \varphi (1)=0] \text{or }\varphi \left(x\right)>0$