When does $2^{n}$ ends with $n$ ?

This method reduces the number of n we have to check.

Note that the unit digits of powers of 2 is $2,4,8,6,2,4...$

n must be even since all powers of 2 are even.

Case 1: $n \equiv 0 \pmod {4}$

The corresponding unit digit for $2^n$ is 6.

n which satisfy both $0 \pmod {4}$ and $6 \pmod 10$ are $n = 16,36,56,\ldots$

Case 2: $n \equiv 2 \pmod {4}$

The corresponding unit digit for $2^n$ is 4.

n which satisfy both $2 \pmod {4}$ and $4 \pmod {10}$ are $n = 14,34,54,\ldots$

Now comes the tedious (but still fairly simple) part - listing the last two digits of powers of 2.

They are $02,04,08,16,32,64,28,56,12,24,48,96,92,84,68,36,72,44,88,76,52,04,08,\ldots$

Hence $2^n \pmod {100}$ has period 20. Checking $14, 16, 34, 36,\ldots$ , we see that their last 2 digits are $84,36,84,36,\dots$

We can stop here since we found the smallest solution of $n=\boxed {36}$

I hope this method is short enough.

${ 2 }^{ 36 }=687194767\boxed { 36 }$

This doesn't work for power with single digits (you may try)

So we must try for two digits. Two digits of 2^n form the following sequence:

2,4,8,16,32,64,28,56,12,24,48,96,92,84,68,36,72,44,88,76,52,4,8,16,32,64...

Now you can try for these values of n (not by hand and human brain only).

You will get the answer n=36