$211!$ or $106^{211}$:Which is greater?

Use a trick similar to Gauss' famous trick for summation (although he probably wasn't the first, if at all, to discover it).

Divide the $211!$ into pairs, $(1,211),(2,210),(3,209),\ldots(105,107)$. Now we can note that for every such pair can be written as $(106-k,106+k)$ and therefore: $$(106-k)(106+k)=106^2-k^2<106^2.$$

Now it's easy to finish. There are $105$ pairs and one additional $106$ to each side. Therefore $211!<106^{211}$.

Apply GM $\le$ AM to $1,\ldots,211$, we have

$$211! \le \left[\frac{1}{211}\left(1 + \ldots + 211\right)\right]^{211} = 106^{211}$$

Since the list $1,\ldots,211$ is not a constant list, the above inequality is strict. i.e. $$211! < 106^{211}$$

$1\cdot 211<106\cdot106$

$2\cdot210<106\cdot106$

$...$

$105\cdot107<106\cdot106$

So clearly $211! < 106^{211}$

Hint: Use Stirling's formula, $n!\simeq\bigg(\!\!\dfrac ne\!\!\bigg)^n\sqrt{2\pi n}$.

Just look at $211!=(106-105)(106-104)...(106-1)106(106+1)...(106+104)(106+105)$ then compare $106^2$ and $(106+n)(106-n)$ when $1\le n \le 105$.