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$(1+a)(1+b)(1+c)(1+d) \\ = 1 + a+b+c+d +ab+ac+ad+bc+bd+cd+abc +abd+acd+bcd+abcd$

Using the AM-GM inequality, we have:

$\small 1 + a+b+c+d +ab+ac+ad+bc+bd+cd+abc +abd+acd+bcd+abcd \ge 16 \sqrt{abcd} = 16$

Equality is established when $a=b=c=d=1$

$\implies (1+a)(1+b)(1+c)(1+d) \ge \boxed{16}$

By AM-GM: $1+a \geq 2 \sqrt{a \times 1}=2 \sqrt{a}$ and similarly for $b,c,d$ :

$(1+a)(1+b)(1+c)(1+d) \geq 16 \sqrt{abcd}=\boxed{16}$