Find the Prime Factorization of $\varphi(11!)$ 
Find the Prime Factorization of $\varphi(11!)$

What I did:
$\varphi(11!)=\varphi(11)\cdot\varphi(10)...\varphi(1)$
$$\varphi(11)=2\cdot 5\\
\varphi(10)=2^2\\
\varphi(9)=3\cdot 2\\
\varphi(8)=2^2\\
\varphi(7)=2\cdot 3\\
\varphi(6)=2\\
\varphi(5)=2^2\\
\varphi(4)=2\\
\varphi(3)=2\\
$$
$$\Longrightarrow\text{ the answer is }  2^{12}\cdot 5\cdot 3^2$$
$11!$
$\varphi(11!)=\varphi(39916800)=8294400$
$8294400=2^{12}3^45^2$
Where am I wrong?
Is there a simpler way to solve this?
 A: \begin{align}
  & \bigg\lfloor \frac{11}{2} \bigg\rfloor+\bigg\lfloor \frac{11}{4} \bigg\rfloor+\bigg\lfloor \frac{11}{8} \bigg\rfloor=8 \\ 
 & \bigg\lfloor \frac{11}{3} \bigg\rfloor+\bigg\lfloor \frac{11}{9} \bigg\rfloor=4 \\ 
 & \bigg\lfloor \frac{11}{5} \bigg\rfloor=2 \\ 
 & \bigg\lfloor \frac{11}{7} \bigg\rfloor=1 \\ 
 &\bigg\lfloor \frac{11}{11} \bigg\rfloor=1 \\ 
\end{align}
$$\varphi(11!)=\varphi ({{2}^{8}}\times {{3}^{4}}\times {{5}^{2}}\times {{7}^{1}}\times {{11}^{1}})=\varphi ({{2}^{8}})\varphi ({{3}^{4}})\varphi ({{5}^{2}})\varphi ({{7}^{1}})\varphi ({{11}^{1}})$$
Note if $P$ be prime then
$$\varphi(P^n)=P^n-P^{n-1}$$
A: While it is true that you should exploit the fact that $\varphi$ is multiplicative, what you do is not quite correct.
Recall that $\varphi(nm)= \varphi(n) \varphi(m)$ needs that $m$ and $n$ are co-prime. 
Thus, first write $11!$ as a product of distinct prime powers. Only, then do what you did. 
A: $\varphi(11!)=(11!) (1-\frac{1}{2})(1-\frac{1}{3}) (1-\frac{1}{5}) (1-\frac{1}{7}) (1-\frac{1}{11})$. I do not really know if this will help.
