bash - Getting errors trying to run shell script (Linux)

I'm trying to run the following script. I have run it before without issue, but now I encounter an error.

#!/bin/bash
# init
function pause(){
   read -p "$*"
}

   echo & echo "(Website):" &&read input
   output=$(ping -c 1 "$input" 2>/dev/null)
   if [ $? -eq 0 ]; then
   ip=$(printf '%s' "$output" | gawk -F'[()]' '/PING/{print $2}'  2>/dev/null)
   clear
   echo
   echo "$ip"; 
   echo -n $ip | xclip -selection c
   echo
   echo
   echo IP copied to clipboard.
   echo
   echo && sleep 2
   pause 'Press [Enter] key to exit...'
   exit
else
   clear
   echo
   echo "Host not found";
   echo && sleep 1
   pause 'Press [Enter] key to exit...'
   exit
fi

But now all of a sudden it results in the following error:

3: /home/username/Desktop/shell.sh: Syntax error: "(" unexpected

I have not made any changes to the script itself, and I don't understand why this bracket results in an error now. Does anybody see anything in this script that I have missed?

Bash version: 4.2.45(1)-release
fresh install, mint 16 (updated everything)

EDIT: I'll put this here so I can actually show what happens.
I ended up removing the pause function completely:

#!/bin/bash
# init

   echo & echo "(Website):" && read input
   output=$(ping -c 1 "$input" 2>/dev/null)
   if [ $? -eq 0 ]; then
   ip=$(printf '%s' "$output" | gawk -F'[()]' '/PING/{print $2}'  2>/dev/null)
   clear
   echo
   echo "$ip"; 
   echo -n $ip | xclip -selection c
   echo
   echo
   echo IP copied to clipboard.
   echo
   echo && sleep 2
   exit
else
   clear
   echo
   echo "Host not found";
   echo && sleep 1
   exit
fi  

But when I run "sh ~/Desktop/shell.sh" I get this:

: not found/Desktop/shell.sh: 3: /home/username/Desktop/shell.sh:  
(Website):  

(I type google.com and press enter) and it returns

: bad variable name/shell.sh: 4: read: 
: not found/Desktop/shell.sh: 5: /home/username/Desktop/shell.sh: 
/home/username/Desktop/shell.sh: 24: /home/username/Desktop/shell.sh: Syntax error: word unexpected

Thank you all for the replies, is there something wrong with the shell itself?

I take back one of the previous version of your script. It's not needed to erase the function.

When I run the script with \bin\bash MyScript.sh seems to function properly.
If instead I run it with \bin\sh MyScript.sh it give me errors.

Why?
In many distributions \bin\sh is a symlink to dash, in other to bash, in other can be the original sh.

If you change with chown u+x MyScript.sh the permission of the script and after you will execute it with ./MyScript.sh the system will use the first line to understand with which shell execute it. You can read more about the shebang e.g. here

If instead you run the script with sh MyScript.sh you force the script to be executed with a different shell (in this case /bin/sh).

#!/bin/bash  

function pause(){  
   read -p "$*"  
}  

printf "\n(Website):" &&read input
output=$(ping -c 1 "$input" 2>/dev/null)
if [ $? -eq 0 ]; then
  ip=$(printf '%s' "$output" | gawk -F'[()]' '/PING/{print $2}'  2>/dev/null)
  clear                                              
  printf "\n${ip}";   
  echo -n $ip | xclip -selection c            
  printf "\n\nIP copied to clipboard.\n\n\n"
  sleep 2 
  pause 'Press [Enter] key to exit...'  
  exit 0  
else
  clear 
  printf "\n\n Host not found\n\n" 
  sleep 1  
  pause 'Press [Enter] key to exit...'  
  exit 1 
fi             

Note:
I change many echo with printf, because in this case it is more compact.
With printf you have to remember to put \n to have a newline, \t for a tab and so on...

Remove the two parentheses () from the definition of your function, so it looks like this:

function pause {
    …
}

And this error:

bad variable name/hey.sh: 4: read: Syntax error: word unexpected

Is because of this:

&&read input

That should be like this:

&& read input

Just tried this with bash 4.2.47-2 on fedora linux 20 64 bits.