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 $10^{100}$
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 $100!>10^{100}$.
Well, if $100! > 10^{100}$ then $10^{100!}$ is bigger. If $10^{100} > 100!$ then a googolplex is bigger.
SO which is bigger $100!$ or a googol?
$100! = (1*2*....*10)*(11*...*20)*(21*....*30)*......*(91*....100)$
$> (1*1*....*1)*(10*10*....*10)*(20*20*...*20)*......*(90*90*...*90)$
$= 1^{10}\times 10^{10}\times 20^{10}\times.... \times 90^{10}$
$=(10^{10})\times (2^{10}*10^{10})\times.... \times (9^{10}*10^{10})$
$= (10^{10}*10^{10}*10^{10}*....*10^{10})\times (2^{10}*3^{10}*.....*9^{10})$
$=(10^{90})\times (2^{10}*3^{10}*4^{10}*5^{10}......*9^{10})$
$> (10^{90})\times (2^{10}*5^{10})$
$= 10^{90}\times 10^{10}$
$= 10^{100}$.
So $100! > 10^{100}$ and
$10^{100!} > 10^{10^{100}}$
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Nother way of thinking of it:
$100!$ has $100$ "pieces" from $1$ to $100$. And $10^{100}$ has $100$ pieces all equal to $10$. The $9$ pieces $1$ to $9$ are all less than $10$ and $90$ of the pieces $11... 100$ are all larger than $10$.
So the question is: Do the $90$ pieces larger than $10$ "overwhelm" the product so $100! > 10^{100}$; or do the $9$ pieces of googol that are $10$ overwhelm the pieces of $100!$ that are less than $10$.
Another way of puttng this is:
$(1*2*....*9)*10*(11*.....*100) <,=,> (10*10*...*10)*10*(10*....*10)\iff$
$\frac {11*.....*100}{10*10*....*10} <,=,> \frac{10*10*...*10}{1*2*3*....*9}$.
Now my intuition says the $1,2,.....9$ are so insignificant and few compared to the many $11,...., 100$.
On the google side we have everything having a geometric average of $10$.
To "make up" for how small the $1$ is, we can pair it with the $100$ so to get $1*100 = 10*10$. And now the $1$ has been "smoothed out".
We can pair the $2$ with the $50$ to get $2*50 =10*10$ and that has been evened out.
We can't pair the $3$ with $33\frac 13$ but if we pair it with $34$ we get $3*34 > 10*10$ so weighting is to $100!$ advantage.
And so on... pair $4*25=10*10$ andd $5*20 =10*10$ and $6*17 > 10*10$ and $7*15 > 10*10$ and $8*13 > 10*10$ and $9*12 > 10*10$.
.... to put this together....
$(1*100)*(2*50)*(3*34)*(4*25)....(9*12) > 100^9 = 10^{18}$.
$10*11*14*16*18*19 > 10^6$
$21*22*23*24*26*27*28*29*30 >10^9$
$31*32*33*35*....*39*40> 10^9$
$41*....*49>10^9$
$51*.....*99>10^{49}$ and so $1*2*....*100 > 10^{18+6+9+9+9+49}=10^{100}$.
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.
You can even check this with google:
$100!>10^{100}$
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.)
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 $\log_{10}(100!) \approx 157.97$)
The numbers from $20$ through $29$ are all $\geq20$, the numbers from $30$ through $39$ are all $\geq30$, and so on, so $$100!>20^{10}\cdot30^{10}\cdot\cdots\cdot90^{10}=(2\cdot3\cdot\cdots\cdot9)^{10}\cdot10^{80}>100^{10}\cdot10^{80}=10^{100}$$
