\fStep 1: To prove that (3 is transcendental, We will use the Lindemann-Weierstrass theorem, Which states that if ay, as, ..., a,, are distinct algebraic numbers, then e*', e, ..., e"" are linearly independent over the field of algebraic numbers. This implies that e\" is transcendental if 3 is algebraic and nonzero. Step 2: Rewrite 3 in terms of an exponential expression: =~ 1 18 - 1071\" n=1 00 1 n" =2 (o) 00 1 n" JB = Z; (6111{10)) B Z (eln(w))\" n=1 JB Z efnn 111{10). n=1 Step 3: If we let o, = n" In(10), then each ,, is an algebraic number since In(10) is transcendental and multiplication or exponentiation by algebraic numbers preserves algebraicity. n" In(10 Applying the Lindemann-Weierstrass theorem, e" = e ) are linearly independent over the field of algebraic numbers. n" In(10 Hence, e ) are linearly independent over the field of algebraic numbers. This implies that 3 is transcendental since e\" would be algebraically independent over the field of algebraic numbers, which means 3 cannot be algebraic. Thus, 3 is transcendental