Notorious Huge

$\begin{aligned} BIG & = & 1 \times 22 \times 333 \times \ldots \times \underbrace{9 \ldots 9}_{ \text{nine} \space 9\text{'s} } \\ & = & \left ( \frac {1}{9} \times 9 \right ) \times \left ( \frac {2}{9} \times 99 \right ) \times \left ( \frac {3}{9} \times 999 \right ) \times \ldots \times \left ( \underbrace{9 \ldots 9}_{ \text{nine} \space 9\text{'s} } \right ) \\ & = & \frac {9!}{9^9} \left ( 10^1 - 1 \right ) \left ( 10^2 - 1 \right ) \left ( 10^3 - 1 \right ) \cdot \cdot \cdot \left ( 10^9 - 1 \right ) \\ & = & \frac {8!}{9^8} 10^{1+2+3+ \ldots + 9} \left ( 1 - \frac {1}{10^1} \right ) \left ( 1 - \frac {1}{10^2} \right ) \left ( 1 - \frac {1}{10^3} \right ) \cdot \cdot \cdot \left ( 1 - \frac {1}{10^9} \right ) \\ & = & \frac {8!}{9^8} 10^{45} \left ( 1 - \frac {1}{10^1} \right ) \left ( 1 - \frac {1}{10^2} \right ) \left ( 1 - \frac {1}{10^3} \right ) \cdot \cdot \cdot \left ( 1 - \frac {1}{10^9} \right ) \\ \end{aligned}$

Because $\left ( 1 - \frac {1}{10^1} \right ) = 0.9$ , $\left ( 1 - \frac {1}{10^2} \right ) = 0.99$ , and the following terms $\approx 1$ , the product of the nine terms is approximately $0.9 \times 0.99$

Continue:

$\begin{aligned} BIG & \approx & \frac {8!}{9^8} 10^{45} (0.9) (0.99) \\ & = & \frac {8!}{9^6} \cdot 10^{44} \cdot 0.11 \\ & = & \frac {8!}{9^5} \cdot 10^{44} \cdot \frac {0.11}{9} \\ & \approx & \frac {8!}{9^5} \cdot 10^{44} \cdot 10^{-2} \\ & = & \frac {8!}{9^5} \cdot 10^{42} \\ \end{aligned}$

Because $\frac {1}{2} < \frac {8!}{9^5} < 1$

$BIG = 10^{42 - \alpha}$ , for a certain fractional part $\frac {1}{2} < \alpha < 1$

Thus, $n = \boxed{42}$

We can rewrite the number $BIG$ into $9! (1 \times 11 \times (\cdots) \times 111111111)$ .

Let's approximate every so-called group to a power of ten:

$9! = 362880 \approx 3.63 \times 10^5$

$111111111*11=1222222221 \approx 1.2 \times 10^9$ $11111111*111=1233333321 \approx 1.2 \times 10^9$ $1111111*1111=1234444321 \approx 1.2 \times 10^9$ $111111*11111=1234554321 \approx 1.2 \times 10^9$

The product of all these five groups is $\approx 3.63 \times (1.2)^4 \times 10^{41} \approx 7.52 \times 10^{41}$ , thus belonging in the interval $[10^{41}, 10^{42}]$ , which means $\boxed{n=42.}$