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The bialgebra pairing condition implies $(1,b_1b_2)=(\Delta(1),b_1\otimes b_2)=(1,b_1)(1,b_2)$, so in particular $(1,1)=(1,1)^2$, and $(1,1)$ is zero or one. Also by a similar calculation, $(b_1b_2,1)=(b_1,1)(b_2,1)$.

We have $$(x,1)=(x,1\cdot1)=(\Delta(x),1\otimes 1)=(x\otimes 1+1\otimes x,1\otimes1)=(x,1)(1,1)+(1,x)(1,1).$$ If $(1,1)=0$ this implies that $(x,1)=0$, and if $(1,1)=1$, this implies that $(1,x)=0$. Together with a similar evaluation of $(1,x)$ this means that $(x,1)=(1,x)=0$. Since $(1,b_1b_2)=(1,b_1)(1,b_2)$ and $(b_1b_2,1)=(b_1,1)(b_2,1)$, we have $(1,x^n)=(x^n,1)=0$ for $n>0$.

We have $$ 1=(x,x)=(x,x\cdot 1)=(\Delta(x),x\otimes 1)=(x\otimes 1+1\otimes x,x\otimes 1)=(x,x)(1,1)+(1,x)(x,1)=(1,1), $$ so $(1,1)=1$.

Clearly, we have $$ (x^m,x^n)=(\Delta(x^m),x\cdot x^{n-1})=\left(\sum_{j=0}^m\binom{m}{j}x^j\otimes x^{m-j},x\otimes x^{n-1}\right). $$

Let us show by induction that $(x^m,x^n)=0$ for $k\ne n$. We already know it for $(m,n)=(0,1),(1,0)$. Suppose it is established for $m+n<N$, where $N\ge 2$. Let us consider $m+n=N$. We may assume $m,n>0$ since $(1,x^k)=(x^k,1)=0$. By the formula above we have $$ (x^m,x^n)=\sum_{j=0}^m\binom{m}{j}(x^j\otimes x^{m-j},x\otimes x^{n-1})=\sum_{j=0}^m\binom{m}{j}(x^j,x)(x^{m-j},x^{n-1}). $$ If $j=0$ or $j=m$, we have $(x^j,x)(x^{m-j},x^{n-1})=0$ since $(1,x^k)=(x^k,1)=0$. Suppose that $0<j<m$. In this case $j+1<m+n$, hence if $j\ne 1$ we have $(x^j,x)=0$ by induction, and if $j=1$, we have $(x^{m-1},x^{n-1})=0$ by induction as $m-1+n-1<m+n$.

Finally, by the same formula for $m=n$ $$ (x^n,x^n)=\sum_{j=0}^n\binom{n}{j}(x^j\otimes x^{m-j},x\otimes x^{n-1})=\sum_{j=0}^n\binom{n}{j}(x^j,x)(x^{n-j},x^{n-1}), $$ and by the result we just proved only $j=1$ has a non-zero contribution, so $$ (x^n,x^n)=n(x^{n-1},x^{n-1}), $$ which immediately gives $$ (x^n,x^n)=n! $$