= Solution
Let $T_q\varphi(x)=\sum_{k=0}^q\varphi^{(k)}(0)x^k/k!$ denote the <Taylor polynomial> at zero. On $0<|x|<1$, the integrand defining the <Hadamard finite-part reciprocal-power distribution> is
$$
x^{-m}\bigl(\varphi(x)-T_{m-2}\varphi(x)\bigr)
=\frac{\varphi^{(m-1)}(0)}{(m-1)!}\frac1x
+x^{-m}\bigl(\varphi(x)-T_{m-1}\varphi(x)\bigr).
$$
The first term is odd, so it cancels under symmetric truncation. The <Taylor remainder> in the second term has absolute value at most $|x|^m\|\varphi^{(m)}\|_\infty/m!$. The integral near zero therefore has a finite limit. For $|x|\geq1$, the original terms are integrable: $x^{-m}\varphi$ has <compact support>, and each subtracted power has exponent $k-m\leq-2$.
This proves existence and linearity. It also gives the global estimate
$$
\begin{aligned}
|\langle\Lambda_m,\varphi\rangle|\leq{}&\frac2{m!}\|\varphi^{(m)}\|_\infty
+\frac2{m-1}\|\varphi\|_\infty\\
&+\sum_{k=0}^{m-2}\frac2{k!(m-k-1)}|\varphi^{(k)}(0)|.
\end{aligned}
$$
Thus
$$
\boxed{\Lambda_m\in\mathcal D'(\mathbb R),\qquad
\operatorname{ord}\Lambda_m\leq m.}
$$
In particular the <order of a distribution> is bounded by one integer on all <compact sets>, not merely separately on each of them.
For the recurrence, write the truncated <Hadamard finite-part integral> as
$$
A_m(\varepsilon,\varphi)=\int_{|x|>\varepsilon}x^{-m}\varphi(x)\,dx
-C_m(\varepsilon,\varphi),
$$
where parity evaluates all subtraction integrals explicitly:
$$
C_m(\varepsilon,\varphi)=
2\sum_{\substack{0\leq k\leq m-2\\m-k\text{ even}}}
\frac{\varphi^{(k)}(0)}{k!(m-k-1)}\varepsilon^{k-m+1}.
$$
Set $C_1=0$ and $\Lambda_1=\operatorname{pv}(1/x)$. On the two truncated intervals, <integration by parts> gives
$$
\int_{|x|>\varepsilon}x^{-m}\varphi'(x)\,dx
=m\int_{|x|>\varepsilon}x^{-m-1}\varphi(x)\,dx
+\varepsilon^{-m}\bigl[(-1)^m\varphi(-\varepsilon)-\varphi(\varepsilon)\bigr].
$$
There are no boundary contributions at infinity. The coefficients of the subtraction terms satisfy
$$
mC_{m+1}(\varepsilon,\varphi)-C_m(\varepsilon,\varphi')
=2\sum_{\substack{0\leq k\leq m-1\\m-k\text{ odd}}}
\frac{\varphi^{(k)}(0)}{k!}\varepsilon^{k-m}.
$$
This is exactly the negative of the boundary term's <Taylor polynomial> through degree $m$; the degree-$m$ term itself vanishes by parity. Hence
$$
\begin{aligned}
&A_m(\varepsilon,\varphi')-mA_{m+1}(\varepsilon,\varphi)\\
&\quad=\varepsilon^{-m}\left[
(-1)^m\bigl(\varphi(-\varepsilon)-T_m\varphi(-\varepsilon)\bigr)
-\bigl(\varphi(\varepsilon)-T_m\varphi(\varepsilon)\bigr)\right]
=O(\varepsilon).
\end{aligned}
$$
Taking the limit proves the full <distributional identity>
$$
\boxed{\langle\Lambda_m,\varphi'\rangle=m\langle\Lambda_{m+1},\varphi\rangle,
\qquad \Lambda_m'=-m\Lambda_{m+1}.}
$$
The same calculation includes $m=1$. Starting with the <distributional derivative of the logarithmic modulus> from part (i), induction yields
$$
\boxed{\Lambda_m=\frac{(-1)^{m-1}}{(m-1)!}\left(\frac d{dx}\right)^m\log|x|,
\qquad c_m=\frac{(-1)^{m-1}}{(m-1)!}.}
$$
This identity holds on all of $\mathbb R$ as a <distributional identity>. It is stronger than matching the ordinary derivatives away from zero, which would leave possible differentiated <Dirac delta distributions> at zero undetermined.
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