The binomial product sum
 \begin{align*}
\sum\limits_{\substack{i_1> i_2> \cdots > i_k\\i_1, i_2, \cdots, i_k \in \{1, 2, \cdots, n-1\}}}\binom{n}{i_1}\binom{i_1}{i_2}\binom{i_2}{i_3}\cdots\binom{i_{k-1}}{i_k}\binom{i_k}{0}
\end{align*}
  with the convention that if the summation is over an empty subset, then only $\binom{n}{0}$ occurs in the product. This formula gives the sequence of Fubini numbers on varying $n$ along  $1,2,3,\ldots$. A justification for this appears in the page https://mathoverflow.net/questions/407239/closed-form-for-a-binomial-product-sum. Now I have a slightly modified question. The same binomial product sum with the following modification
 \begin{align*}
\sum\limits_{\substack{i_1> i_2> \cdots > i_k\\i_1, i_2, \cdots, i_k \in \{1, 2, \cdots, n-1\}}}(-1)^{n+k+1}\binom{n}{i_1}\binom{i_1}{i_2}\binom{i_2}{i_3}\cdots\binom{i_{k-1}}{i_k}\binom{i_k}{0} 
\end{align*} gives the constant sequence $1,1,1,\ldots$ for $n=1,2,3,\ldots$ as verified using the sagemath code posted [here][1]. The cancellations in the summation is not very insightful if one check some concrete example. Is there any way to say that this sum is indeed going to give 1 all the times? What could be a justification for this?


  [1]: https://sagecell.sagemath.org/?z=eJyNU8tqwzAQvBf6D4tzkWIlxKX0UPA_FNpbcI3jrFPZjiRkudC_r-RHJCelRAex7GN2d0Y6YgWnXFa5yuu8lFhVRLGavj4-gD2rGlJ4Hu2ibfOuP3TW0_LOkHdro-nIXm3qOGHufmLbrTWSjIHNPPet4TYl_dA9UjohKs2FITPY5B3vSmooBjca4OLScRrGnTGYl4U-ujlQkLmA-qTZte2kNkTjN-oOxyl8Eq9IAJamOxq08TPNx0gz0mPbkk1CP4ndk64PXMgzL1pLGlgGKKyglFpjp6Q4cnGydWC-EPCszA_YbkvUUgrDRY_eOwDmSku3X4g-L7XfZRTWPnTxB9s4Bdw0HtZbswC3tHnLCdE4BfY7ljhNQ-yn7IqoayaTayaHvTQWzX8EB4uv0z_Wa7KAgyZOMrqoX4JNS0ZvFq4vDUTsLpzFDJPK8eKRxUmQ7t9EnAa193A9RXzRFNNoei3IBdm6nRhqECNhwwd7mQUYMW5-r6IMUBzTiEX0F0ZhEyY=&lang=sage&interacts=eJyLjgUAARUAuQ==