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We note that $3!^{10} = 6^{10} \equiv 1 \pmod{11}$ . We also note that since $5!$ is a multiple of $10$ , $5!^n$ is also a multiple of $10$ , therefore,

$3!^{5!^{7!^{...}}} \equiv 6^{10} \pmod {11} \equiv \boxed{1} \pmod{11}$

Similarly, $3!^{5!} = 6^{120} \equiv 1 \pmod{13} \quad \Rightarrow 3!^{5!^{7!^{...}}} \equiv 1 \pmod{13}$

3! = 6 then 6^5 mod 11 = -1 mod 11 or 10 mod 11 so it becomes (-1)^ even or 1mod 11 = 1

$3! = 6$

$6^1 = 6 \equiv 6$ mod $11$

$6^2 = 36 \equiv 3$ mod $11$

$6^3 \equiv 18 \equiv 7$ mod $11$

$6^5 \equiv 3*7 = 21 \equiv -1$ mod $11$

$6^{10} \equiv (-1)^2 = 1$ mod $11$ .

Now the exponent of $3! = 6$ is a (LARGE) power of $5!$ that is clearly a multiple of $5$ and $2$ , hence a multiple of $10$ . So

$\Large3!^{5!^{7!^{.^{.^{.^{2013!}}}}}} = 6^{(10n)^m}= 6^{10^m n^m} =$ $\Large = \left((6^{10})^{10^{m-1}}\right)^{n^m} \equiv \left((1)^{10^{m-1}}\right)^{n^m} = 1 \quad \textrm{ mod }11$

We know that 3!^5!^7!....^2013! = (3!)^(5!.7!....2013!) = (3)^(5!.7!....2013!) *(2)^(5!.7!....2013!) Now the trick is to first find the power of 3 that when divided by 11 leaves maximum possible or least possible reminder. Similarly for 2. In this case it is 10 or 1. We see 3^5 = 243 when divided by 11 leaves a reminder 1 and 2^5 = 32 when divided by 11 leaves a reminder 10 (which can be taken as -1) .

So [(3)^(5!.7!....2013!) *(2)^(5!.7!....2013!)]/11= [(3^5)^(4!.7!....2013!)/11 *(2^5)^(4!.7!....2013!)/11]

Per reminder theorem, The reminder is: (1)^(4!7!...2013!)*(-1)^4!7!...2013!) Since 4!.7!....2013! is always even (since it ends with zero) , the remainder is 1