Compute the standard factorization of $15!$.
Solve:
\begin{equation}
\sum_{r=1}^{\infty}[\frac{15}{2^r}]=11
\end{equation}
\begin{equation}
\sum_{r=1}^{\infty}[\frac{15}{3^r}]=6
\end{equation}
\begin{equation}
\sum_{r=1}^{\infty}[\frac{15}{5^r}]=3
\end{equation}
\begin{equation}
\sum_{r=1}^{\infty}[\frac{15}{7^r}]=2
\end{equation}
\begin{equation}
\sum_{r=1}^{\infty}[\frac{15}{11^r}]=1
\end{equation}
\begin{equation}
\sum_{r=1}^{\infty}[\frac{15}{13^r}]=1
\end{equation}
So according to Elementary methods in number theory Theorem 1.12 vp(n!)=∑[logpn]r=1[npr] ,
\begin{equation}
15!=2^{11}\times 3^6\times 5^3\times 7^2\times 11\times 13
\end{equation}