A random pattern or an insane function

Let f(z) be a function defined on z such that it forms the palindrome of z. $Eg:f(1234)=4321$ $$ We have, $\text{1st term}=f(1^{10})=1$ $\text{2nd term}=f(2^{10})=4201$ $\text{3rd term}=f(3^{10})=94095$ $.$ $.$ $.$ $\text{99th term}=f(99^{10})$

From Binomial theorem,on expansion of 99th term,it is found that it doesn't has 6 $$ => x = 6 $\text{6th term}= f(6^{10})$ Hence,Palindrome of 6th term is $f(f(6^{10})) = 6^{10}= 60466176$

Brilliant problem. The word palindrome gave it away. Or else I would never have figured out the sequence. Reversing the digits of the numbers in the given sequence, we get $1,1024,59049$ Observe that they are all 10th powers of natural numbers $1^{10},2^{10},3^{10}$ Since the 99th term is to be found, use wolframalpha to find the value $99^{10} = 904382075008804490001$ Observe that the digit 6 is missing. So the 6th term will be the reverse of $6^{10}$ But the question asks us to find the palindrome of the 6th term, which will simply be $6^{10} = \boxed{60466176}$

Confession: I used oeis.org to identify the sequence. (I'd still be peering at this problem tomorrow if I hadn't.)

Here f calculates terms in the sequence in the form of strings rather than integers. I verify that the digit 6 is not in the string for $f(99)$ . Then I calculate $f(6)$ and its value with digits reversed.