Lucky 7

$7^{77}=(7^{11)^7}>7^7\color{#D61F06}{(as 7^{11}>7)}$

The quickest way is to change the exponents to 7, which can be achieved by rewriting $7^{77}$ into $(7^{11})^7$ . As $77 < 7^{11}$ , We can then conclude that $77^{7} < (7^{11})^7$ , or simply $77^7 < 7^{77}$ . Hence, the statement is FALSE

This problem could be solved by using intuition:

Let's look at how many digits would $7^{77}$ and $77^{7}$ have.

We can notice that in $7^{x}$ , $x$ increases a number by one digit $\frac{6}{7}$ times it goes up by $1$ (What I mean is that when $x$ is $1$ , the number has $1$ digit, when $x$ is $2$ - $2$ digits and so on and on until $x = 7$ , then it will have only $6$ digits). That means that $77-\frac{77}{7}=66$ digits will be in a number (for the sake of safety we will say that the least digits that would be in $7^{77}$ is $50$ . It's mainly for time sake).

Now, even if we will think that in $77^{x}$ , $x$ always increases a number by $2$ digits, we would get maximum of $14$ digits.

I think that we all know the truth of "The more digits a number has, the bigger it is". So, when we compare a number with at least $50$ digits ( $7^{77}$ ) with a number that has maximum of $14$ digits ( $77^{7}$ ), we clearly see that $\boxed{7^{77}}$ is a lot bigger.

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Please note, that this solution is based merely on logical thinking and has very little "real" math behind it. It is a way to go around the problem, to solve it faster.