The Conway knot has normalized Alexander polynomial one

Let $V$ be the matrix in [F1] and form $A(t)=V-t{V}^T$. Fraction-free Bareiss elimination uses, at stage $k$, the update $a_{ij}^{(k+1)}=\frac{a_{kk}^{(k)}{a}_{ij}^{(k)}-a_{ik}^{(k)}{a}_{kj}^{(k)}}{a_{k-1,k-1}^{(k-1)}}(i,j>k),$ with initial divisor $1$. The successive pivots are $t-1,1-t+t^2,-t+t^2,t^2,t-t^2+t^3-t^4,t-2t^2+3t^3-2t^4+t^5\mathrm{.}$ At the seventh pivot exchange rows $7$ and $8$ once; the pivot is $t^2-t^3+t^4$ and the final Bareiss entry is $-t^4$. The one row exchange reverses the determinant sign, so $\det A(t)=t^4$. The recurrence has exact polynomial divisions at every stage; the finite pivot and exact-division certificate is recorded in the batch contract.

The monomial $t^4$ is a unit in $\mathbb{Z}[t,t^{-1}]$. By [F2], the Alexander generator is associated to $1$. The polynomial $1$ is symmetric and takes value $1$ at $t=1$, so uniqueness in [F3] gives ${\mathrm{\Delta }}_C^{\mathrm{\#}}(t)=1$.