To simplify the expression \( 78 \cdot 77^8 \cdot 7^{7} \), we will first rewrite \( 78 \) in terms of its prime factorization:
\[ 78 = 2 \cdot 3 \cdot 13 \]
Next, let's look at \( 77 \):
\[ 77 = 7 \cdot 11 \]
Now substitute the factorization of \( 77 \) into the original expression:
\[ 77^8 = (7 \cdot 11)^8 = 7^8 \cdot 11^8 \]
Now, putting this back into the expression, we have:
\[ 78 \cdot 77^8 \cdot 7^7 = 78 \cdot (7^8 \cdot 11^8) \cdot 7^7 \]
Combine all the \( 7 \) terms:
\[ = 78 \cdot 7^{8+7} \cdot 11^8 = 78 \cdot 7^{15} \cdot 11^8 \]
Now, replace \( 78 \) with its factorization:
\[ = (2 \cdot 3 \cdot 13) \cdot 7^{15} \cdot 11^8 \]
Since there are no further simplifications possible with the factors, we can write the final result as:
\[ 2 \cdot 3 \cdot 13 \cdot 7^{15} \cdot 11^8 \]
Thus, the expression in exponential form is:
\[ 2^1 \cdot 3^1 \cdot 13^1 \cdot 7^{15} \cdot 11^8 \]
If needed, it can also be presented as:
\[ 78 \cdot 7^{15} \cdot 11^8 \]
That would be the most simplified form of the given expression.