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

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.

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