Compute the sum of sum of sum of digits of $1976^{1976}$

I have this problem:

Compute $$sd(sd(sd(1976^{1976})))$$, where $$sd()$$ represents the sum of digits.

I know that $$1976 = 2^3 \cdot 19 \cdot 13$$, so we have to compute $$sd(sd(sd((2^3 \cdot 19 \cdot 13)^{1976})))$$.

I also know that $$sd(a\cdot b) = sd(sd(a)\cdot sd(b))$$, so I can rewrite the above as: $$sd(sd(sd( sd(2^{3\cdot1976}) \cdot sd(13^{1976}) \cdot sd(19^{1976})))$$ but I think this is not the right direction.

I also know that $$sd(m\cdot n) \leq sd(m)\cdot sd(n)$$, so theoretically I should be able, maybe, to approximate the sum, but how to compute the sum of sum of sum of digits of that number ($$1976^{1976})$$?

Thank you very much!

Note: I previously asked how to compute $$sd(1976^{1976})$$ but it seems there was a typo in my book and the real question is the question I ask here. I edited the old question, to the iterative sum and accepted the correct answer.

• @JoséCarlosSantos I do think the problem is slightly different. – Don Thousand Oct 22 '18 at 15:42
• According to WA the sum of digits of $1976^{1976}$ is $29239$ and the sum of digits of that is (obviously) $25$ so the next one up is $7$. This suggests: try to prove that your value is just the iterated sum of digits. – lulu Oct 22 '18 at 15:43

It is easy to see that the iterated sum of digits of your expression is $$7$$, thus your answer must be congruent to $$7 \pmod 9$$.
Note that $$\log_{10}{1976^{1976}}\approx 6512.474994$$ so your expression has $$6513$$ digits. Thus the sum of digits of your expression is not greater than $$9\times 6513=58617$$
Now, that has $$5$$ digits so the second sum of digits is less than $$9\times 5=45$$. Now, inspection quickly shows that there is no number $$≤45$$ which has a sum of digits greater than $$12$$. Thus the answer to your question is also $$7$$.
You need to calculate $$1976^{1976}\mod {9}$$. Be decomposing $$1976$$ as you did we have $$1976^{1976}=(2^3.19.13)^{1976}=(4.19.26)^{1976}$$as we know $$26\cong -1\mod 9\\19\cong 1\mod 9$$therefore$$1976^{1976}\cong 2^{2\times 1976}\mod 9\cong 2^{2\times 1974}\cdot 16\cong 7\mod 9$$so the sum of the digits after too many levels is $$7$$