Is there an easy formula for $\operatorname{Tor}_i^{\mathbb{Z}/(p^n)}(\mathbb{Z}/(p),\mathbb{Z}/(p))$?

I've been following some old slides from the Spring 2010 Algebra Seminar at UWaterloo.

I now know that $$\operatorname{Ext}_{\mathbb{Z}/(p^n)}^i(\mathbb{Z}/(p),\mathbb{Z}/(p))\cong\mathbb{Z}/(p)$$ for all $i\geq 0$ when $n>1$. Otherwise, when $n=1$, every $\operatorname{Ext}$ group vanishes for $i\geq 1$.

However the Tor groups aren't addressed as closely in the examples, so I would like to know about the other side/left side of the coin. Is there a simple formula or characterization of $\operatorname{Tor}_i^{\mathbb{Z}/(p^n)}(\mathbb{Z}/(p),\mathbb{Z}/(p))$ for $i\geq 0$ for $n$ arbitrary?

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Let $\newcommand\ZZ{\mathbb Z}R=\ZZ/p^n$, with $n>1$, be your ring and consider the module $M=\ZZ/p$. The obvious map $R\to M$ has kernel the ideal generated by $p$, so the sequence $R\xrightarrow{p}R\to M$ is exact. The kernel of the map $p:R\to R$ is generated by $p^{n-1}\in R$. We thus have an exact sequence of the form $$R\xrightarrow{\quad p^{n-1}\quad} R\xrightarrow{\quad p\quad}R\to M$$ Now the kernel of $p^{n-1}:R\to R$ is precisely the ideal generated by $p$, so we can extend the exact sequence to $$R\xrightarrow{\quad p\quad}R\xrightarrow{\quad p^{n-1}\quad} R\xrightarrow{\quad p\quad}R\to M$$ We can continue in this way, and conclude that there is a projective resolution of $M$ as an $R$-module which is of infinite length and with maps $p$ and $p^{n-1}$, repeaing with period $2$: $$\cdots \to R\xrightarrow{\quad p\quad}R\xrightarrow{\quad p^{n-1}\quad} R\xrightarrow{\quad p\quad}R\xrightarrow{\quad p^{n-1}\quad} R\xrightarrow{\quad p\quad}R\to M$$ To compute $Tor$, we drop the $M$ on the right, apply the functor $(-)\otimes_R M$ and compute the homology of the resulting complex. If you do that, you will find that the differentials in the complex are all zero, and all the modules isomorphic to $M$: it follows that $Tor_i(M,M)\cong M$ for all $i\geq0$.