Which is bigger: a googolplex or $10^{100!}$ A googol is defined as $ 10^{100}$
Let x = $10^{100}$
A googolplex is defined as $10^{x}$
Which is bigger: a googolplex or $10^{100!}$
I only know that:
$100! = 1×2×3×...×98×99×100$
$10^{100} = 10×10×10×...×10×10×10$
I think its easier to approach if I only compare the exponents, because they both have the same base $10$ anyways, but I don't know how to show which is bigger from $100!$ and $10^{100}$
 A: Both $10^{100}$ and $100!$ have 100 terms.
Observe that
$$(101-k)×k=101k-k^2\geq 100=10^2$$
for $k=1,...,50$ and equality holds only for $k=1$.
Thus you can put togheter 2 by 2 the terms in $100!$ so that their product la bigger than $10^2$. Thus $100!>10^{100}$.
A: The question boils down to whether $10^{100}$ is greater than $100!$ (same bases). But by Stirling's approximation,
$$100!\approx\sqrt{200\pi}(100/e)^{100}$$
and $100/e>33>10$ and $\sqrt{200\pi}<30$ so $100!>10^{100}$. Hence $10^{100!}$ is the bigger number.
A: You can even check this with google:
$100!>10^{100}$
A: By regrouping terms in $100!$, using some crude inequalities, and carefully counting the number of terms involved, we have
$$\begin{align}
100!
&=100\cdot99\cdot98\cdots11\cdot10\cdot9\cdots2\cdot1\\
&=(100\cdot1)(99\cdot2)\cdots(92\cdot9)\times(91\cdot90\cdots11\cdot10)\\
&\gt(100)(100)\cdots(100)\times(10\cdot10\cdots10\cdot10)\\
&=100^9\cdot10^{82}\\
&=10^{100}\\
&=\text{googol}
\end{align}$$
Therefore $10^{100!}\gt10^\text{googol}=\text{googolplex}$
(Remark: The "$\times$" symbol's role here is purely visual, to put a little extra separation between things that are treated differently. The answer, in general, is very similar to Alberto Saracco's.)
A: Consider that
$$
\log_{10}(100!) = \sum_{i=1}^{100}\log_{10}(i) \ge \sum_{i=32}^{100}\log_{10}(i)\ge \sum_{i=32}^{100}\frac{3}{2} = 103.5 > 100
$$
Now exponentiate both sides twice to get
$$
10^{100!} > 10^{10^{100}}
$$
(Note that if you want a better estimate for the size of $10^{100!}$, the logarithmic sum can be computed exactly fairly easily, giving $\log_{10}(100!) \approx 157.97$)
A: The numbers from $20$ through $29$ are all $\geq20$, the numbers from $30$ through $39$ are all $\geq30$, and so on, so 
$$100!>20^{10}\cdot30^{10}\cdot\cdots\cdot90^{10}=(2\cdot3\cdot\cdots\cdot9)^{10}\cdot10^{80}>100^{10}\cdot10^{80}=10^{100}$$
