Tensor Algebra: Symmetrization & Antisymmetrization Problem
Given the tensor algebra:
$$TV:=\sum_{k=0}^\infty{\bigotimes}^k V$$
Regard the symmetrization and antisymmetrization:
$$S_\pm\left(\bigotimes_{i=1}^kv_i\right):=\frac{1}{k!}\sum_{\sigma\in\mathcal{S}_k}(\pm1)^{\mathrm{sgn}\sigma}\bigotimes_{i=1}^kv_{\sigma(i)}$$
Then they satisfies:
$$S\pm\left(S_\pm\left(\bigotimes_{i=1}^kv_i\right)\otimes v_{k+1}\right)=S_\pm\left(\bigotimes_{i=1}^kv_i\otimes v_{k+1}\right)$$
How to prove these relations?
Attempt
After changing summations they become:
$$\frac{1}{k+1!}\frac{1}{k!}\sum_{\sigma\in\mathcal{S}_{k}}\sum_{\tau\in\mathcal{S}_{k+1}}(\pm1)^{\mathrm{sgn}\sigma+\mathrm{sgn}\tau}\bigotimes_{i=1}^kv_{\tau(\sigma(i))}\otimes v_{\tau(k+1)}$$
Substituting inner permutation gives:
$$\tau'_\sigma(1\leq i\leq k):=\tau(\sigma(i))\quad\tau'_\sigma(k+1):=\tau:\quad\mathrm{sgn}\tau'_\sigma=\mathrm{sgn}\sigma+\mathrm{sgn}\tau$$
Then the outer sum repeats so they reduce to the desired:
$$\frac{1}{k+1!}\sum_{\tau'_\sigma\in\mathcal{S}_{k+1}}(\pm1)^{\mathrm{sgn}\tau'_\sigma}\bigotimes_{i=1}^{k+1}v_{\tau'_\sigma(i)}$$
But was to deal with the signum right?

There is no problem with this proof, but there is one point that requires a bit of clarification. Note that if we change the index of summation to $\tau_{\sigma}'$ we get
$$\frac{1}{(k+1)!}\frac{1}{k!}\sum_{\sigma\in S_k}{\sum_{\tau_{\sigma}'\in S_{k+1}}{(\pm 1)^{\mathrm{sgn}(\tau_{\sigma}')}\bigotimes_{i=1}^{k+1}{v_{\tau'_{\sigma}(i)}}}}$$
Although ostensibly this depends on $\sigma$, all the application of $\sigma$ does is permute the order of the terms. The sign of the permutation is correct, so regardless of what $\sigma$ is the sum is the same. Thus we are summing the same thing $k!$ times, hence the result is
$$\frac{1}{(k+1)!}\sum_{\tau_{\sigma}'\in S_{k+1}}{(\pm 1)^{\mathrm{sgn}(\tau_{\sigma}')}\bigotimes_{i=1}^{k+1}{v_{\tau'_{\sigma}(i)}}}$$
