This will probably go over your head OBT but this is how expansions work for powers.
\((a+b)^n=\binom{n}{0}*a^nb^0+\binom{n}{1}*a^{n-1}b+\binom{n}{2}*a^{n-2}b^2+\dots \binom{n}{n}*a^{0}b^{n}\\~\\ now\;\;\binom{n}{0},\;\binom{n}{1},\binom{n}{3},\;\dots \binom{n}{n}\\ \qquad \qquad \text{Is the nth row of pascal's triangle (if the first row is taken as 1,1)} \\\\~\\ \text{Lets see what hapens if a=10 and b=1}\\ (10+1)^n=\binom{n}{0}*10^n*1^0+\binom{n}{1}*10^{n-1}*1+\binom{n}{2}*10^{n-2}*1^2+\dots \binom{n}{n}*10^{0}*1^{n}\\ (10+1)^n=\binom{n}{0}*10^n+\binom{n}{1}*10^{n-1}+\binom{n}{2}*10^{n-2}*+\dots \binom{n}{n}*10^{0}\\ (10+1)^n=\binom{n}{0}*10^n+\binom{n}{1}*10^{n-1}+\binom{n}{2}*10^{n-2}*+\dots \binom{n}{n}*10^{0}\\ (10+1)^n=1*10^n+\binom{n}{1}*10^{n-1}+\binom{n}{2}*10^{n-2}*+\dots 1*10^{0}\\\)
Now I can see that the digits of this number will give pascals triangle BUT only while every nCr term is a single digit.
In other words it will only work up to 11^4 = 14641
I'll look at 11^4 and try and explain:
\( 11^4=(10+1)^n\\ =1*10^4+\binom{4}{1}*10^{3}+\binom{4}{2}*10^{2}*+\binom{4}{3}*10^{1}+\binom{4}{4}*10^{0}\\ =1*10^4+\binom{4}{1}*10^{3}+\binom{4}{2}*10^{2}*+\binom{4}{3}*10+\binom{4}{4}\\ =1*10000+\binom{4}{1}*1000+\binom{4}{2}*100+\binom{4}{3}*10+\binom{4}{4}\\ =1*10000+4*1000+6*100+4*10+1\\ =14641\)
It will not work for 11^5 becasue there will be carry, try it and you will find out. :)
