Ever thought of that the binomial koefficients equal 11<sup>n</sup>?

Ever thought of that the binomial koefficients equal 11<sup>n</sup>?
The sum of binomial coefficients is equal to 2<sup>n</sup>. For example, <sup>5</sup>C<sub>0</sub> + <sup>5</sup>C<sub>1</sub> + <sup>5</sup>C<sub>2</sub> + <sup>5</sup>C<sub>3</sub> + <sup>5</sup>C<sub>4</sub> + <sup>5</sup>C<sub>5</sub> = 1 + 5 + 10 + 10 + 5 + 1 = 32 = 2<sup>5</sup>.
Proof: Take x = 1 in the binomial expansion of (1+x)<sup>n</sup>.
Yeah, I thought that's perhaps what he meant.
I think he meant that the digits of 11^n are binomial coefficients.
1
11
121
1331
14641
161051 < It breaks here though
101^n lasts a while longer. In general, any 10^k+1 works for a while and the reason is pretty simple. (a + b)^n = sum over i of (n choose i)*(a^i)*(b^(ni)), so for a=10, b=1 you're just left with powers of 10 times the binomial coefficients.
You are right about what I meant Magi Master but the only reason that the binomial coefficients aren't palindroms all the time even if they equal 11<sup>n</sup> is that ten symbols, as we use, aren't enough.
There will be overflow.
If you want to you can write a computer program with for example 15 symbols and there you'll see that 11<sup>5</sup> is a palindrom.
Yes, in a higher base, 11<sup>n</sup> will stay a palindrome for longer, but never indefinitely (101<sup>n</sup> is about the same as 11<sup>n</sup> in base100). I suppose you could define something like a baseinfinity (based on tuples) that would stay a palindrome but I don't think there's much point in that (correct me if I'm wrong).
The only point is that palindromes are beautiful.
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