Let $(X,\mathcal{M},\mu)$ be a measurable space, $1<p_{k}<\infty$ for $k=1,2,\ldots,N$ with $\sum_{k=1}^{N}\frac{1}{p_{k}}=1$ and $f_{k}\in L^{p_{k}}(\mu)$ for $k=1,\ldots,N$. Show that
$$\left\|f_{1}f_{2}\cdots f_{N}\right\|_{1}\leq \|f_{1}\|_{p_{1}}\|f_{2}\|_{p_{2}}\cdots\|f_{N}\|_{p_{N}}.$$
Remark: I tried to use the Holder inequality with $p=p_{1} $ and $q=\frac{1}{\sum_{k=2}^{\infty}\frac{1}{p_{k}}}$, but when I try to continue the process, it becomes complicated until I think that this is not the way.
$\|f_{1}\cdots f_{N}\|_{1}\leq\|f_{1}\|_{p_{1}}\|f_{2}\cdots f_{N}\|_{r}$, where $r$ is such that $1/r=1/p_{2}+\cdots+1/p_{N}$. Now $1=1/(p_{2}/r)+\cdots+1/(p_{N}/r)$ and $\|f_{2}\cdots f_{N}\|_{r}^{r}=\|f_{2}^{r}\cdots f_{N}^{r}\|_{1}\leq\|f_{2}\|_{p_{2}}^{r}\cdots\|f_{N}\|_{p_{N}}^{r}$ by induction.
Note that, for example, $\|f_{2}^{r}\|_{p_{2}/r}=\left\{\displaystyle\int\left(|f_{2}|^{r}\right)^{p_{2}/r}\right\}^{r/p_{2}}=\left[\left\{\displaystyle\int|f_{2}|^{p_{2}}\right\}^{1/p_{2}}\right]^{r}=\|f_{2}\|_{p_{2}}^{r}$.