Here's a more numberical proof:
Let x = infinity
1^infinity = 1^(x), take the ln of 1^(x) (mathematically legal as long as you raise the answer to e in the end), you get
ln(1)^(x) = xln(1) = ln(1) / (1/x)
take the limit
limit x ->infinity : ln(1) / (1/x)
ln1 = 0
1/x = 0
thus, you have 0/0 which is undefined. Therefore, 1^infinity is undefined.
when you play with infinity, algebraic reasoning doesnt apply. you need calculus tools.
Let x = infinity
1^infinity = 1^(x), take the ln of 1^(x) (mathematically legal as long as you raise the answer to e in the end), you get
ln(1)^(x) = xln(1) = ln(1) / (1/x)
take the limit
limit x ->infinity : ln(1) / (1/x)
ln1 = 0
1/x = 0
thus, you have 0/0 which is undefined. Therefore, 1^infinity is undefined.
when you play with infinity, algebraic reasoning doesnt apply. you need calculus tools.