Let $n$ be $m$. It's a consequence of Generalized Hölder inequality: if $f_i \in L^{p_i}$ for all $i = 1 , \ldots , n$, where $p_i \in (1 , \infty)$ for all $i = 1 , \ldots , n$ and $\sum_{i = 1}^n \frac{1}{p_i} = 1$, then $f_1 \cdot \ldots \cdot f_n \in L^1$ and
$$
{\|f_1 \cdot \ldots \cdot f_n\|}_1 \leq {\|f_1\|}_{p_1} \cdot \ldots \cdot {\|f_n\|}_{p_n}\mbox{.}
$$
The proof can do also for $p_i \in [1 , + \infty]$, $i = 1 , \ldots , n$, but here we suppose that $1 \neq p_i \neq + \infty$ for all $i = 1 , \ldots , n$.
Proof. We do that first supposing that $f_i$ is nonnegative for all $i = 1 , \ldots , n$. Let us reason by induction: si $n = 2$, the inequality is true by Hölder inequality. So we suppose the hypothesis is true for $n \in \mathbb{N}$, with $n > 2$, and let's show it for $n + 1$. Let $f_{n + 1} : X \to \mathbb{R}$ be a measurable nonnegative function and let ${\{q_i\}}_{i = 1}^{n + 1}$ be a finite sequence of real numbers, with $q_i \in (1 , \infty)$ for all $i = 1 , \ldots , n , n + 1$, such that $\sum_{i = 1}^{n + 1} \frac{1}{q_i} = 1$. We consider $\tilde{f_n} = f_n f_{n + 1}$ and $\tilde{q_n} = \frac{q_n q_{n + 1}}{q_n + q_{n + 1}}$, because of this
$$
\frac{1}{\tilde{q_n}} = \frac{1}{\frac{q_n q_{n + 1}}{q_n + q_{n + 1}}} = \frac{q_n + q_{n + 1}}{q_n q_{n + 1}} = \frac{q_n}{q_n q_{n + 1}} + \frac{q_{n + 1}}{q_n q_{n + 1}} = \frac{1}{q_{n + 1}} + \frac{1}{q_n} = \frac{1}{q_n} + \frac{1}{q_{n + 1}}\mbox{.}
$$
So it's clair that $\tilde{f_n}$ is a measurable nonnegative function and $\tilde{q_n} \in (1 , \infty)$ because, how we have supposed that exists $j \in \{1 , \ldots , n - 1\}$ such that $q_j \neq 0$,
$$
\frac{1}{\tilde{q_n}} = \frac{1}{q_n} + \frac{1}{q_{n + 1}} = 1 - \sum_{i = 1}^{n - 1} \frac{1}{q_i} < 1 \Longleftrightarrow \tilde{q_n} > 1\mbox{.}
$$
Then, using induction hypothesis,
$$
\int_X (f_1 \cdot \ldots \cdot f_n f_{n + 1}) d \mu = \int_X (f_1 \cdot \ldots \cdot f_{n - 1} \tilde{f_n}) d \mu \leq
$$
$$
\leq {\left(\int_X f_1^{q_1} d \mu\right)}^{\frac{1}{q_1}} \cdot \ldots \cdot {\left(\int_X f_{n - 1}^{q_{n - 1}} d \mu\right)}^{\frac{1}{q_{n - 1}}} {\left(\int_X \tilde{f_n}^{\tilde{q_n}} d \mu\right)}^{\frac{1}{\tilde{q_n}}}\mbox{,}
$$
that is to say
\begin{equation}
\int_X (f_1 \cdot \ldots \cdot f_n f_{n + 1}) d \mu \leq {\left(\int_X f_1^{q_1} d \mu\right)}^{\frac{1}{q_1}} \cdot \ldots \cdot {\left(\int_X f_{n - 1}^{q_{n - 1}} d \mu\right)}^{\frac{1}{q_{n - 1}}} {\left(\int_X \tilde{f_n}^{\tilde{q_n}} d \mu\right)}^{\frac{1}{\tilde{q_n}}}\mbox{.} \tag{1}
\end{equation}
If we prove that
$$
{\left(\int_X \tilde{f_n}^{\tilde{q_n}} d \mu\right)}^{\frac{1}{\tilde{q_n}}} \leq {\left(\int_X f_n^{q_n} d \mu\right)}^{\frac{1}{q_n}} {\left(\int_X f_{n + 1}^{q_{n + 1}} d \mu\right)}^{\frac{1}{q_{n + 1}}}\mbox{,}
$$
we will have finished the proof. We consider $q_n^* = \frac{q_{n + 1}}{q_n + q_{n + 1}}$. It's clair, becasue $q_n , q_{n + 1} > 1$, that $q_n^* \in (0 , 1)$. We consider $p = \frac{q_n + q_{n + 1}}{q_{n + 1}}$. Then $p = \frac{1}{q_n^*} \in (1 , \infty)$ and his conjugated exponent is $p' = \frac{q_n + q_{n + 1}}{q_n}$; in fact,
$$
\frac{1}{p} + \frac{1}{p'} = \frac{q_{n + 1}}{q_n + q_{n + 1}} + \frac{q_n}{q_n + q_{n + 1}} = \frac{q_{n + 1} + q_n}{q_n + q_{n + 1}} = \frac{q_n + q_{n + 1}}{q_n + q_{n + 1}} = 1\mbox{.}
$$
How $\tilde{q_n} \in (1 , \infty)$ and $f_n$ and $f_{n + 1}$ are measurables and nonnegatives, then $f_n^{\tilde{q_n}}$ and $f_{n + 1}^{\tilde{q_n}}$ are measurables and nonnegatives also and, using Hölder inequality for nonnegative functions,
$$
\int_X \tilde{f_n}^{\tilde{q_n}} d \mu = \int_X {(f_n f_{n + 1})}^{\tilde{q_n}} d \mu = \int_X f_n^{\tilde{q_n}} f_{n + 1}^{\tilde{q_n}} d \mu \leq
$$
$$
\leq {\left(\int_X {(f_n^{\tilde{q_n}})}^p d \mu\right)}^{\frac{1}{p}} {\left(\int_X {(f_{n + 1}^{\tilde{q_n}})}^{p'} d \mu\right)}^{\frac{1}{p'}} = {\left(\int_X f_n^{\tilde{q_n} p} d \mu\right)}^{\frac{1}{p}} {\left(\int_X f_{n + 1}^{\tilde{q_n} p'} d \mu\right)}^{\frac{1}{p'}}\mbox{,}
$$
that is to say
\begin{equation}
\int_X \tilde{f_n}^{\tilde{q_n}} d \mu \leq {\left(\int_X f_n^{\tilde{q_n} p} d \mu\right)}^{\frac{1}{p}} {\left(\int_X f_{n + 1}^{\tilde{q_n} p'} d \mu\right)}^{\frac{1}{p'}}\mbox{.} \tag{2}
\end{equation}
On the one hand
\begin{equation}
\tilde{q_n} p = \frac{q_n q_{n + 1}}{q_n + q_{n + 1}} \frac{q_n + q_{n + 1}}{q_{n + 1}} = \frac{q_n q_{n + 1} (q_n + q_{n + 1})}{q_{n + 1} (q_n + q_{n + 1})} = q_n\mbox{.} \tag{3}
\end{equation}
On the other hand,
\begin{equation}
\tilde{q_n} p' = \frac{q_n q_{n + 1}}{q_n + q_{n + 1}} \frac{q_n + q_{n + 1}}{q_n} = \frac{q_n q_{n + 1} (q_n + q_{n + 1})}{q_n (q_n + q_{n + 1})} = q_{n + 1}\mbox{.} \tag{4}
\end{equation}
\indent Then, uniting $(2)$, $(3)$ and $(4)$,
$$
\int_X \tilde{f_n}^{\tilde{q_n}} d \mu \leq {\left(\int_X f_n^{\tilde{q_n} p} d \mu\right)}^{\frac{1}{p}} {\left(\int_X f_{n + 1}^{\tilde{q_n} p'} d \mu\right)}^{\frac{1}{p'}} = {\left(\int_X f_n^{q_n} d \mu\right)}^{\frac{1}{p}} {\left(\int_X f_{n + 1}^{q_{n + 1}} d \mu\right)}^{\frac{1}{p'}}
$$
and, composing with $t \mapsto t^{\frac{1}{\tilde{q_n}}}$ (it's growing on $[0 , \infty)$), we obtain, using again $(3)$ and $(4)$,
$$
{\left(\int_X \tilde{f_n}^{\tilde{q_n}} d \mu\right)}^{\frac{1}{\tilde{q_n}}} \leq {\left({\left(\int_X f_n^{q_n} d \mu\right)}^{\frac{1}{p}} {\left(\int_X f_{n + 1}^{q_{n + 1}} d \mu\right)}^{\frac{1}{p'}}\right)}^{\frac{1}{\tilde{q_n}}} = {\left(\int_X f_n^{q_n} d \mu\right)}^{\frac{1}{p \tilde{q_n}}} {\left(\int_X f_{n + 1}^{q_{n + 1}} d \mu\right)}^{\frac{1}{p' \tilde{q_n}}} =
$$
$$
= {\left(\int_X f_n^{q_n} d \mu\right)}^{\frac{1}{\tilde{q_n} p}} {\left(\int_X f_{n + 1}^{q_{n + 1}} d \mu\right)}^{\frac{1}{\tilde{q_n} p'}} = {\left(\int_X f_n^{q_n} d \mu\right)}^{\frac{1}{q_n}} {\left(\int_X f_{n + 1}^{q_{n + 1}} d \mu\right)}^{\frac{1}{q_{n + 1}}}\mbox{,}
$$
that is to say
\begin{equation}
{\left(\int_X \tilde{f_n}^{\tilde{q_n}} d \mu\right)}^{\frac{1}{\tilde{q_n}}} \leq {\left(\int_X f_n^{q_n} d \mu\right)}^{\frac{1}{q_n}} {\left(\int_X f_{n + 1}^{q_{n + 1}} d \mu\right)}^{\frac{1}{q_{n + 1}}}\mbox{.} \tag{5}
\end{equation}
\indent Definitely, uniting $(1)$ and $(5)$,
$$
\int_X (f_1 \cdot \ldots \cdot f_n f_{n + 1}) d \mu \leq {\left(\int_X f_1^{q_1} d \mu\right)}^{\frac{1}{q_1}} \cdot \ldots \cdot {\left(\int_X f_{n - 1}^{q_{n - 1}} d \mu\right)}^{\frac{1}{q_{n - 1}}} {\left(\int_X \tilde{f_n}^{\tilde{q_n}} d \mu\right)}^{\frac{1}{\tilde{q_n}}} \leq
$$
$$
\leq {\left(\int_X f_1^{q_1} d \mu\right)}^{\frac{1}{q_1}} \cdot \ldots \cdot {\left(\int_X f_{n - 1}^{q_{n - 1}} d \mu\right)}^{\frac{1}{q_{n - 1}}} {\left(\int_X f_n^{q_n} d \mu\right)}^{\frac{1}{q_n}} {\left(\int_X f_{n + 1}^{q_{n + 1}} d \mu\right)}^{\frac{1}{q_{n + 1}}}\mbox{,}
$$
that is to say
$$
\int_X (f_1 \cdot \ldots \cdot f_n f_{n + 1}) d \mu \leq {\left(\int_X f_1^{q_1} d \mu\right)}^{\frac{1}{q_1}} \cdot \ldots \cdot {\left(\int_X f_n^{q_n} d \mu\right)}^{\frac{1}{q_n}} {\left(\int_X f_{n + 1}^{q_{n + 1}} d \mu\right)}^{\frac{1}{q_{n + 1}}}\mbox{.}
$$
Let's suppose now that $f_i \in L^{p_i}$ for all $i = 1 , \ldots , n$. Then
$$
\int_X {|f_i|}^{p_i} d \mu < \infty \Longleftrightarrow {\left(\int_X {|f_i|}^{p_i} d \mu\right)}^{\frac{1}{p_i}} < \infty \Longleftrightarrow {\|f_i\|}_{p_i} < \infty
$$
for all $i = 1 , \ldots , n$, so
$$
{\|f_1\|}_{p_1} \cdot \ldots \cdot {\|f_n\|}_{p_n} < \infty\mbox{,}
$$
because of this, using Generalized Hölder inequality for nonnegative functions,
$$
{\|f_1 \cdot \ldots \cdot f_n\|}_1 = \int_X |f_1 \cdot \ldots \cdot f_n| d \mu = \int_X (|f_1| \cdot \ldots \cdot |f_n|) d \mu \leq
$$
$$
\leq {\left(\int_X {|f_1|}^{p_1} d \mu\right)}^{\frac{1}{p_1}} \cdot \ldots \cdot {\left(\int_X {|f_n|}^{p_n} d \mu\right)}^{\frac{1}{p_n}} = {\|f_1\|}_{p_1} \cdot \ldots \cdot {\|f_n\|}_{p_n} < \infty
$$
and, in particular,
$$
\int_X |f_1 \cdot \ldots \cdot f_n| d \mu < \infty \Longleftrightarrow f_1 \cdot \ldots \cdot f_n \in L^1\mbox{.}
$$
