Find the last two digits of : `N=(1!+2!+3!+.......+101!)^101`

Find the last two digits of : `N=(1!+2!+3!+.......+101!)^101` 

Note: This is equivalent to finding `N(mod11)`.

ie: The remainder when dividing `N` by `100`.

Observation: `10!\equiv0(mod100)` Because, `10!=10...5...2`

Therefore, `N\equiv(1!+2!+3!+......+9!)^101(mod100)`

                   `N\equiv(1+2+6+24+20+20+40+20+80)^101(mod100)`

                   `N\equiv13^101(mod100)`

As we can see:

                `13^2=169`

                `13^4=..61`

                `13^5=....93`

                `13^20=(13^5)^4=....01`

Then,     `N\equiv13(13^20)^5(mod100)`

Hence, The last two digits of `N` is `13`.