Show that for $n_1,n_2,...,n_p$ natural numbers, there is a natural number $n$ such that $\varphi(n)=n_1!n_2!...n_p!$

Question

Show that for $n_1,n_2,...,n_p$ natural numbers, there is a natural number $n$ such that $\varphi(n)=n_1!n_2!...n_p!$. In the proof, they order the numbers such that $n_1\leq n_2 ...\leq n_p$. Let $p_1,p_2,...,p_m$ prime numbers smaller or equal to $n_p$. They are the sole prime numbers in decomposing $N=n_1!n_2!...n_p!$. I don't understand why they say that, if $p_i\leq n_j$, then every $p_i$ is a factor of $n_j!$ and why $(p_i-1)$ divides $N$, such that $N=(p_1-1)(p_2-1)...(p_m-1)p_1^{k_1}p_2^{k_2}...p_m^{k_m}$ and then $p_1-1,p_2-1,...,p_m-1$ can be further decomposed using this factors and so $N=\varphi(n)$, where$n=p_1^{k_1+1}p_2^{k_2+1}...p_m^{k_m+1}$.