[proofplan] We reduce modulo a maximal ideal. Since $M$ is nonzero and finitely generated, Nakayama's Lemma provides a maximal ideal $\mathfrak{m}$ with $M / \mathfrak{m}M \neq 0$. The surjection $M \twoheadrightarrow M / \mathfrak{m}M$ induces, by right-exactness of the tensor product applied $n - 1$ times, a surjection $M^{\otimes n} \twoheadrightarrow (M / \mathfrak{m}M)^{\otimes n}$. The target is a tensor power of a nonzero finite-dimensional vector space over the field $R / \mathfrak{m}$, which has positive dimension, hence is nonzero. Therefore $M^{\otimes n} \neq 0$. [/proofplan] [step:Find a maximal ideal $\mathfrak{m}$ with $M / \mathfrak{m}M \neq 0$ via Nakayama's Lemma] Since $M \neq 0$, the annihilator $\operatorname{Ann}_R(M) = \{r \in R : rM = 0\}$ is a proper ideal of $R$ (as $1_R \notin \operatorname{Ann}_R(M)$). By Zorn's Lemma, there exists a maximal ideal $\mathfrak{m} \trianglelefteq R$ containing $\operatorname{Ann}_R(M)$. We claim $M / \mathfrak{m}M \neq 0$. Suppose for contradiction that $\mathfrak{m}M = M$. Consider the localisation $M_\mathfrak{m}$ at $\mathfrak{m}$. Since $M$ is finitely generated over $R$, the localisation $M_\mathfrak{m}$ is finitely generated over the local ring $(R_\mathfrak{m}, \mathfrak{m}R_\mathfrak{m})$. The assumption $\mathfrak{m}M = M$ implies $\mathfrak{m}R_\mathfrak{m} \cdot M_\mathfrak{m} = M_\mathfrak{m}$, so by [Nakayama's Lemma](/theorems/???) applied in the local ring $R_\mathfrak{m}$, we get $M_\mathfrak{m} = 0$. However, since $\operatorname{Ann}_R(M) \subseteq \mathfrak{m}$, the localisation $M_\mathfrak{m} \neq 0$: if $M_\mathfrak{m} = 0$, then for each generator $m_i$ of $M$ (say $M$ is generated by $m_1, \ldots, m_k$) there exists $s_i \in R \setminus \mathfrak{m}$ with $s_i m_i = 0$. The product $s = s_1 \cdots s_k \in R \setminus \mathfrak{m}$ (since $\mathfrak{m}$ is prime) satisfies $s m_i = 0$ for all $i$, so $s \in \operatorname{Ann}_R(M) \subseteq \mathfrak{m}$, a contradiction. Therefore $\mathfrak{m}M \neq M$, i.e., $M / \mathfrak{m}M \neq 0$. [guided] The goal is to find a maximal ideal $\mathfrak{m}$ for which the "reduction modulo $\mathfrak{m}$" of $M$ is nonzero. The quotient $M / \mathfrak{m}M$ is naturally a vector space over the residue field $k = R / \mathfrak{m}$ (the $R$-module structure factors through $R / \mathfrak{m}$ since $\mathfrak{m}$ acts as zero). Not every maximal ideal works: it is possible that $\mathfrak{m}M = M$ for some $\mathfrak{m}$. The condition we need is $\operatorname{Ann}_R(M) \subseteq \mathfrak{m}$, which ensures $M$ is "genuinely supported" at $\mathfrak{m}$. Since $M \neq 0$, the annihilator $\operatorname{Ann}_R(M)$ is a proper ideal ($1_R$ does not annihilate $M$), so it is contained in some maximal ideal $\mathfrak{m}$. We claim $M / \mathfrak{m}M \neq 0$ for this $\mathfrak{m}$. Suppose for contradiction that $\mathfrak{m}M = M$. Localise at $\mathfrak{m}$: $M_\mathfrak{m}$ is a finitely generated module over the local ring $(R_\mathfrak{m}, \mathfrak{m}R_\mathfrak{m})$ satisfying $\mathfrak{m}R_\mathfrak{m} \cdot M_\mathfrak{m} = M_\mathfrak{m}$. [Nakayama's Lemma](/theorems/???) gives $M_\mathfrak{m} = 0$. But $\operatorname{Ann}_R(M) \subseteq \mathfrak{m}$ prevents this. Write $M = Rm_1 + \cdots + Rm_k$. If $M_\mathfrak{m} = 0$, then for each $i$ there exists $s_i \in R \setminus \mathfrak{m}$ with $s_i m_i = 0$. The product $s = s_1 \cdots s_k$ lies in $R \setminus \mathfrak{m}$ (since $\mathfrak{m}$ is prime, a product of elements outside $\mathfrak{m}$ stays outside $\mathfrak{m}$) and satisfies $s M = 0$, i.e., $s \in \operatorname{Ann}_R(M) \subseteq \mathfrak{m}$, contradicting $s \notin \mathfrak{m}$. Therefore $M / \mathfrak{m}M$ is a nonzero vector space over $k = R / \mathfrak{m}$. [/guided] [/step] [step:Construct a surjection $M^{\otimes n} \twoheadrightarrow (M / \mathfrak{m}M)^{\otimes n}$ using right-exactness] Let $\pi: M \twoheadrightarrow M / \mathfrak{m}M$ denote the canonical surjection. Since $\pi$ is surjective and the tensor product functor is right-exact, the $R$-module homomorphism \begin{align*} \pi^{\otimes n}: M^{\otimes n} \to (M / \mathfrak{m}M)^{\otimes n}, \quad m_1 \otimes \cdots \otimes m_n \mapsto \pi(m_1) \otimes \cdots \otimes \pi(m_n), \end{align*} is surjective. More precisely, this map is the $n$-fold iteration of the right-exactness property: the surjection $\pi \otimes \operatorname{id}_{M^{\otimes(n-1)}}: M \otimes_R M^{\otimes(n-1)} \twoheadrightarrow (M/\mathfrak{m}M) \otimes_R M^{\otimes(n-1)}$ is surjective by right-exactness, and then $\operatorname{id}_{M/\mathfrak{m}M} \otimes (\pi^{\otimes(n-1)}): (M/\mathfrak{m}M) \otimes_R M^{\otimes(n-1)} \twoheadrightarrow (M/\mathfrak{m}M) \otimes_R (M/\mathfrak{m}M)^{\otimes(n-1)}$ is surjective by induction on the number of tensor factors. Composing these surjections gives the surjection $\pi^{\otimes n}$. [/step] [step:Conclude that $(M / \mathfrak{m}M)^{\otimes n} \neq 0$ since it is a tensor power of a nonzero vector space] The module $\overline{M} := M / \mathfrak{m}M$ is a nonzero vector space over the field $k = R / \mathfrak{m}$, with finite dimension $d := \dim_k(\overline{M}) \geq 1$. The tensor power $\overline{M}^{\otimes n}$ is taken over $R$, but since $\mathfrak{m}$ acts as zero on $\overline{M}$, it factors through the tensor power over $k$: \begin{align*} \overline{M}^{\otimes_R n} \cong \overline{M}^{\otimes_k n}. \end{align*} The right-hand side is a $k$-vector space of dimension $d^n \geq 1$ (if $\{e_1, \ldots, e_d\}$ is a $k$-basis of $\overline{M}$, then $\{e_{i_1} \otimes \cdots \otimes e_{i_n} : 1 \leq i_1, \ldots, i_n \leq d\}$ is a $k$-basis of $\overline{M}^{\otimes_k n}$, which has $d^n$ elements). In particular, $\overline{M}^{\otimes n} \neq 0$. Since $\pi^{\otimes n}: M^{\otimes n} \twoheadrightarrow \overline{M}^{\otimes n}$ is surjective and the target is nonzero, the source $M^{\otimes n}$ must be nonzero. This completes the proof. [guided] The final step uses the key property that distinguishes tensor products over fields from tensor products over general rings: over a field, the tensor product of nonzero vector spaces is always nonzero. This is because vector spaces are free modules, and the tensor product of free modules is free with rank equal to the product of the ranks. Concretely, let $d = \dim_k(\overline{M}) \geq 1$ and let $\{e_1, \ldots, e_d\}$ be a $k$-basis. The tensor power $\overline{M}^{\otimes_k n}$ has basis $\{e_{i_1} \otimes \cdots \otimes e_{i_n}\}$ as $(i_1, \ldots, i_n)$ ranges over $\{1, \ldots, d\}^n$. This basis has $d^n \geq 1$ elements, so $\overline{M}^{\otimes_k n} \cong k^{d^n} \neq 0$. Why does the tensor power over $R$ agree with the tensor power over $k$? Every element of $\mathfrak{m}$ acts as zero on $\overline{M} = M / \mathfrak{m}M$, so the $R$-bilinear map $\overline{M} \times \overline{M} \to \overline{M} \otimes_R \overline{M}$ factors through the $k$-bilinear map, giving $\overline{M} \otimes_R \overline{M} \cong \overline{M} \otimes_k \overline{M}$. The same applies iteratively to all $n$ factors. Since $\pi^{\otimes n}$ maps $M^{\otimes n}$ surjectively onto the nonzero module $\overline{M}^{\otimes n}$, we conclude $M^{\otimes n} \neq 0$. [/guided] [/step]