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

Let $a=211!$ and $b=106^{211}$. Show which is greater with proper logic.

By matching term by term,it is pretty easy to note that 
$106!<106^{106}$
$106^{105}<107\cdot 108\cdot 109...........211$
However I am at a loss to see how this will help.The solution must something along the lines of this one but I am unable to see so.I also factorized 106 but it complicated matters further.A hint will be appreciated.
 A: Hint: Use Stirling's formula, $n!\simeq\bigg(\!\!\dfrac ne\!\!\bigg)^n\sqrt{2\pi n}$.
A: 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 $1\le n \le 105$.
A: 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 $211!<106^{211}$.
A: 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.
$$211! < 106^{211}$$
A: $1\cdot 211<106\cdot106$
$2\cdot210<106\cdot106$
$...$
$105\cdot107<106\cdot106$
So clearly $211! < 106^{211}$
A: The problem may be "simplified" by taking logarithms of both numbers. Then:
$$\frac{\displaystyle\sum_{k=2}^{211}\log(k)}{211\log(106)}<\frac{922}{984}<1.$$
This implies $$\sum_{k=2}^{211}\log(k)<211\log(106),$$
or in other words
$$211!<106^{211}.$$
EDIT: This still leaves the problem of working out the logarithms in the first inequality. There may be a nice way of proving the first inequality.
A: You have received very good answers to your post. But, let me go further and suppose  the problem to be : find the smallest value of $x$ such that  $$211! < x^{211}$$  Using, as suggested by Lucian, Stirling's approximation of $n!$ and taking logarithms of both sides immediately leads to $921.837 < 211 \space \log(x)$ that is to say $x >78.9564$. So, the answer is 79.   
