Solving $ T' = 0 $ for distributions in $\mathbb{R}^n$

Question

Denoting $ T \in \mathcal{D}'(\mathbb{R}^n) $ as distributions with $ T_f(\varphi) = \int_{\mathbb{R}^n} f\varphi\ dx $, I wish to prove the distribution solution of the equation $ T' = 0 $ (distribution derivative) is $ T = T_c $ where $c$ is constant.
I have a proof for $\mathbb{R}$, where for all $\varphi \in C^\infty_c(\mathbb{R})$ we can have $ \varphi = \psi + \varphi_0 T_1(\varphi) $ where $$ \psi = \varphi - \varphi_0 T_1(\varphi) $$ choosing $ \varphi_0 \in C^\infty_c(\mathbb{R}) $ such that $ T_1(\varphi_0) = \int_\mathbb{R} \varphi_0 dx =1 $. Hence we have $ T_1(\psi)= \int_\mathbb{R}\psi dx = 0 $, so for $ \tau(x) = \int_{-\infty}^x \psi(t)dt $ we have $ \psi = \tau'$. Hence if $ T' = 0 $ then $$ T(\varphi ) = T(\tau') + T(\varphi_0)T_1(\varphi) = T(\varphi_0)T_1(\varphi) = T_{T(\varphi_0)}(\varphi) $$ Thus $ T = T_{T(\varphi_0)} $. But I can't replicate this on $\mathbb{R}^n$. Any proof for $\mathbb{R}^n $ would be extremely helpful.