Упростите выражение: [latex](a+1)(a^{2}+1)(a^{4}+1)(a^{8}+1)(a^{16}+1)(a^{32}+1)[/latex]

по формуле разности квадратов [latex](a-b)(a+b)=a^2-b^2[/latex] если [latex] a\neq 1 [/latex] [latex](a+1)(a^2+1)(a^4+1)(a^8+1)(a^{16}+1)(a^{32}+1)=\\ \frac{(a-1)(a+1)(a^2+1)(a^4+1)(a^8+1)(a^{16}+1)(a^{32}+1)}{a-1}=\\ \frac{(a^2-1)(a^2+1)(a^4+1)(a^8+1)(a^{16}+1)(a^{32}+1)}{a-1}=\\ \frac{(a^4-1)(a^4+1)(a^8+1)(a^{16}+1)(a^{32}+1)}{a-1}=\\ \frac{(a^8-1)(a^8+1)(a^{16}+1)(a^{32}+1)}{a-1}=\\ \frac{(a^{16}-1)(a^{16}+1)(a^{32}+1)}{a-1}=\\ \frac{(a^{32}-1)(a^{32}+1)}{a-1}=\\ \frac{a^{64}-1}{a-1}[/latex] если а=1, то [latex](a+1)(a^2+1)(a^4+1)(a^8+1)(a^{16}+1)(a^{32}+1)=(1+1)(1^2+1)(1^4+1)(1^8+1)(1^{16}+1)(1^{32}+1)=(1+1)(1+1)(1+1)(1+1)(1+1)(1+1)=2*2*2*2*2*2=8*8=64[/latex]