help please

The last digit of a multiplication problem is the last digit of the product of the two unit digits.

First make a list of the unit digits of the sequence and see if there is a pattern.

1, [2, 4, 8, 2, 6, 2], [2, 4, 8, 2, 6, 2], [2, 4, 8, 2, 6, 2]

It looks like after 1 there is a 6 digit repeating pattern [2, 4, 8, 2, 6, 2]

Ignoring the 1 at the beginning of the sequence the problem is now find the 34th term in the pattern that was found.

The largest multiple of 6 under 34 is 30 so the 30th term will be the last 2 in the pattern.

30- 2

31- 2

32- 4

33- 8

34- 2

The unit digit of the 35th term in the original sequence is \(\boxed{2}\)

I just wrote it out, and got 2

This sequence consists of Fibonacci Numbers raised to the power of 2 as follows:

Fibonacci Sequence: 1 , 1 , 2 , 3 , 5 , 8 , 13 , 21 , 34 , 55.......[10 terms]

Your sequence        : 1, 2, 2, 4, 8, 32, 256........... Your 7th term = 2^(Fib 6th term) = 2^8 = 256

Therefore: Your 35th term =2^34F. 34th Fib. term =5,702,887  =2^5,702,887, which will give 8 as the last digit as can be seen here in the last 20 digits:.....49292157765605654528.

In the sequence

1, 2, 2, 4, 8, 32, 256,...,

each term starting from the third term) is the product of the two terms before it. For example, the seventh term is 256, which is the product of the fifth term (8) and the sixth term (32). 

This sequence can be continued forever, though the numbers very quickly grow enormous!

(For example, the 14th term is close to some estimates of the number of particles in the observable universe.)

What is the last digit of the 35th term of the sequence?

see: https://web2.0calc.com/questions/help-please_92099#r8

thanks so much!