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How can we obtain $$\sum_{l=1}^{m}(-1)^{l+1} \binom{m}{l} \equiv 1 \pmod{p} $$

Please give me an idea to obtain this.

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  • $\begingroup$ not sure what you are asking exactly. for which values of $(m,p)$ the equality holds? $\endgroup$ – gt6989b Jan 21 '20 at 21:34
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    $\begingroup$ Idea: forget about $p$ and try to compute $$\sum_{l = 0}^{m} (-1)^l \binom{m}{l}\,.$$ $\endgroup$ – Daniel Fischer Jan 21 '20 at 21:36
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    $\begingroup$ expand $(1-1)^m=0$ using the binomial theorem $\endgroup$ – J. W. Tanner Jan 21 '20 at 21:42
  • $\begingroup$ @J.W.Tanner Thank you so much! $\endgroup$ – Oily Jan 21 '20 at 22:58