219. What is greater, $100^{300}$ or 300!?
219. Let's prove preliminarily that the product of \( n \) consecutive integers is greater than the square root of the product of the extreme numbers raised to the power of \( n \). Let's denote these numbers as \( a, a+1, \ldots, a+n-1 \). Then the \( k \)-th number from the beginning is \( a+k-1 \), and the \( k \)-th number from the end is \( a+n-k \). Their product is

\[
\begin{aligned}
(a+k-1)(a+n-k) &= a^2 + a n - a + (k-1)(n-k) \\
& \geqslant a^2 + a n - a = a(a+n-1)
\end{aligned}
\]

where equality can only be achieved when \( k=1 \) or \( k=n \). In other words, the product of two numbers equidistant from the ends (in the case of an odd \( n \), these two numbers can, in particular, both be equal to the middle number) is always greater than the product of the extreme numbers. Therefore, the product of all numbers

\[
a(a+1) \ldots (a+n-1) \geqslant [a(a+n-1)]^{\frac{n}{2}} = [\sqrt{a(a+n-1)}]^n,
\]

where the equality sign holds only for \( n=1 \) or \( n=2 \).

Now let's prove that \( 3001 > 100^{300} \). We have:

\[
\begin{aligned}
& 1 \cdot 2 \ldots 25 > \sqrt{25^{25}} = 5^{25} ; \\
& 26 \ldots 50 > (\sqrt{26 \cdot 50})^{25} > 35^{25} ;
\end{aligned}
\]

\[
\begin{aligned}
& 51 \ldots 100 > (\sqrt{51 \cdot 100})^{50} > 70^{50} \\
& 101 \ldots 200 > \sqrt{100^{100}} \cdot \sqrt{200^{100}} = 10^{200} \cdot 2^{50} \\
& 201 \ldots 300 > \sqrt{200^{100}} \cdot \sqrt{300^{100}} = 10^{200} \cdot 2^{50} \cdot 3^{50}
\end{aligned}
\]

Multiplying the left and right sides of the inequalities, we get:

\[
\begin{aligned}
& 3001! > 5^{25} \cdot 35^{25} \cdot 70^{50} \cdot 10^{400} \cdot 2^{100} \cdot 3^{50} = \\
&= 5^{50} \cdot 7^{25} \cdot 5^{50} \cdot 14^{50} \cdot 10^{400} \cdot 2^{100} \cdot 3^{50} = 10^{500} \cdot 7^{25} \cdot 14^{50} \cdot 3^{50} = \\
&= 10^{500} \cdot 21^{25} \cdot 42^{25} \cdot 14^{25} > 10^{500} \cdot 20^{25} \cdot 40^{25} \cdot 14^{25} = \\
&= 10^{550} \cdot 2^{25} \cdot 4^{25} \cdot 14^{25} = 10^{550} \cdot 112^{25} = \\
& \quad= 10^{600} \cdot 1.12^{25} > 10^{600} = 100^{300}
\end{aligned}
\]

Note. A more general result is formulated in problem 223.