I have been searching the internet for this one and found a link to the page "1, 2, 3, 4 - four digits that DWARFED the universe".
On the bottom of this page there is a table with a few of the great combinations that can be made with $1$, $2$, $3$, and $4$. If you look for the entries with just the numbers (no decimals, no minus, etc.), you see the following:
- $2^{3^{41}}$ has got $10^{19}$ digits
- $3^{4^{21}}$ has got $10^{12}$ digits
- $3^{421}$ has got $201$ digits
- $4^{321}$ has got $194$ digits
- $2^{431}$ has got $130$ digits
You can calculate/test numbers yourself with this gigantic online calculator: HyperCalc Javascript
(Trying to give an explanation): As @h34 notes in a comment: We need to use exponentiation to get the biggest number. If you draw a graph of an exponential function it shoots upward after a few numbers. For instance the simple binary function $2^x$ results in: $$2, 4, 8, 16, 32, 64, 128, 256, \ldots$$ (we all know this sequence). More precisely:
- Any expression of the type $x + 1$ or $x \cdot 1$ or $x^1$ or $1^x$ can be given a bigger value by juxtaposing the digit $1$ to some other digit in the expression $x$. Therefore the digit $1$ will not stand alone in the expression, but will be juxtaposed to some other digit. Furthermore, the digit $1$ will come after the other digit(s) in this juxtaposition.
- For $x, y \ge 2$, we always have $x^y \ge x+y$ and $x^y \ge x \cdot y$. Therefore, the maximizing expression will only use exponentiations.
- Any expression $x$ without exponentiation will have value at most $4321$.
- Any expression $x^y$ with only a single exponentiation will have value at most $3^{421}$ (which is about $7.38×10^{200}$).
- Therefore it remains to consider expressions of the form $a^{b^{c}}$ where one of $a,b,c$ is $21$, $31$, or $41$, and where the other two are single-digit numbers.
- If $a$ consists of two digits then the value of $a^{b^c}$ is at most $41^{3^2}$,
which is much smaller than $2^{3^{41}}$. Therefore $a$ is a single digit.
If $b$ has two digits and $c\ge2$ is single digit, then $b^c<c^b$. Therefore $b$ is single digit and $c$ has two digits.
Hence, it only remains to analyze the six possible candidates
$2^{3^{41}}$, $2^{4^{31}}$, $3^{2^{41}}$, $3^{4^{21}}$, $4^{2^{31}}$, and $4^{3^{21}}$.
For the fifth candidate, we have $$4^{2^{31}} = 2^{2 \cdot 2^{31}} = 2^{2^{32}},$$
whereas for the sixth candidate, we have $$4^{3^{21}} = 2^{2 \cdot 3^{21}} < 2^{3^{22}}.$$
Hence both candidates are clearly smaller than the first candidate $2^{3^{41}}$.
For the fourth candidate, we have $$3^{4^{21}} = 3^{2^{2 \cdot 21}} = 3^{2^{42}},$$ which dominates and eliminates the third candidate $3^{2^{41}}$.
The fourth candidate is $$3^{4^{21}} < 4^{4^{21}} = 2^{2^{43}}.$$ The last exponent $2^{43}$ satisfies $$2^{43} = 2^6 \cdot 2^{37} < 81 \cdot 3^{37} = 3^{41},$$ and hence is smaller than the exponent of the first candidate. This eliminates the fourth candidate.
Finally, let us compare the exponents of the first and the second candidate:
$$4^{31} = 2^{11} \cdot 2^{51} = 2048 \cdot 8^{17} < 2187 \cdot 9^{17} = 3^7 \cdot 3^{34} = 3^{41}.$$
This eliminates the second candidate.
Summarizing: The answer is that the first candidate $2^{3^{41}}$ yields the largest possible value.
Spoiler
If we want to get the answer as big as we possibly can, we need to use the highest numbers as many times as possible within the calculation. To achieve this we use exponentiation multiple times with the numbers in increasing order: $1^{2^{3^4}}$. The $1$ doesn't do anything in this calculation, so we put it together with the highest number $4$ to create an even bigger number: $2^{3^{41}}$
Another Spoiler
And to answer the question, why not $3^{2^{41}}$? Here is a graph of both $2^x \cdot \log(3) $ and $3^x \cdot \log(2)$. These functions give the best view about the differences between them.
Indeed, $3^{2^x}$ has a head start (begins at $3$ rather than $2$), but it quickly passed by the $2^{3^x}$-function when $x > 1.14$ and never crosses again.
