Prove by induction that the differentiation of distributions is defined by the following integration-by-parts formula: For any distribution $f$ and a “nice” function $\varphi$, we have
$$\int_{\mathbb{R}^{n}} dx \ \partial^{m}f(x) \ \varphi (x) = (-1)^{|m|} \int_{\mathbb{R}^{n}} dx \ f(x) \ \partial^{m}\varphi (x)$$
This is what I have done
$1)$ Base case $m=0$
$$\int_{\mathbb{R}^{n}} dx \ \partial^{0}f(x) \ \varphi (x) = (-1)^{|0|} \int_{\mathbb{R}^{n}} dx \ f(x) \ \partial^{0}\varphi (x) \Rightarrow \int_{\mathbb{R}^{n}} dx \ f(x) \ \varphi (x) = \int_{\mathbb{R}^{n}} dx \ f(x) \ \varphi (x)$$
OK.
$2)$ Induction hypothesis: the statement holds for $m$ (i.e. $\int_{\mathbb{R}^{n}} dx \ \partial^{m}f(x) \ \varphi (x) = (-1)^{|m|} \int_{\mathbb{R}^{n}} dx \ f(x) \ \partial^{m}\varphi (x)$ holds).
$3)$ Induction step: show that the statement holds for $m+1$. The trick is integrating by parts
$$\int_{\mathbb{R}^{n}} dx \ \partial^{m+1}f(x) \ \varphi (x) = \partial^{m}f(x) \ \varphi (x)+(-1)\int_{\mathbb{R}^{n}} dx \ \partial^{m}f(x) \ \partial \varphi (x) $$
Applying integration by parts $m$ times one gets
$$\int_{\mathbb{R}^{n}} dx \ \partial^{m+1}f(x) \ \varphi (x) =$$ $$=\partial^{m}f(x) \ \varphi (x)-\partial^{m-1}f(x) \ \partial \varphi (x) +...+(-1)^{|m+1|}\int_{\mathbb{R}^{n}} dx \ f(x) \ \partial^{m+1} \varphi (x) $$
So we see that to finish our proof we need $\partial^{m}f(x) \ \varphi (x)-\partial^{m-1}f(x) \ \partial \varphi (x) +...$ to vanish. But how can we mathematically justify this? I guess it has to do with assuming boundary conditions. If that is the case, could do please explain me how to do so?
Besides I did not have to use the induction hypothesis. This has never happened to me before while using induction. I guess we need it to show $\partial^{m}f(x) \ \varphi (x)-\partial^{m-1}f(x) \ \partial \varphi (x) +...=0$?